Ch.02 Modeling in Frequency Domain

1/11/2016 System Dynamics and Control 2.01 Modeling in Frequency Domain 02 Modeling in Frequency Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.03 Modeling in Frequency Domain §1.Introduction System Dynamics and Control 2.02 Modeling in Frequency Domain Chapter Objectives After completing this chapter, the student will be able to • Find the Laplace transform of time functions and the inverse Laplace transform • Find the transfer function (TF) from a differential equation and solve the differential equation using the transfer function • Find the transfer function for linear, time-invariant electrical networks • Find the TF for linear, time-invariant translational mechanical systems • Find the TF for linear, time-invariant rotational mechanical systems • Find the TF for gear systems with no loss and for gear systems with loss • Find the TF for linear, time-invariant electromechanical systems • Produce analogous electrical and mechanical circuits • Linearize a nonlinear system in order to find the TF HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.04 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review - Transforms: a mathematical conversion from one way of thinking to another to make a problem easier to solve - Mathematical models from schematics of physical systems • transfer functions in the frequency domain • state equations in the time domain - The Laplace transform the problem in time-domain to problem in 𝑠-domain, then applying the solution in 𝑠-domain, and finally using inverse transform to converse the solution back to the time-domain - The Laplace transform is named in honor of mathematician and astronomer Pierre-Simon Laplace (1749-1827) - Others: Fourier transform, z-transform, wavelet transform, … HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.05 Modeling in Frequency Domain §2.Laplace Transform Review - The Laplace transform of the function 𝑓(𝑡) for 𝑡 > is defined by the following relationship HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.06 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review - The Laplace transform table +∞ 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 𝐹 𝑠 = ℒ 𝑓(𝑡) = (2.1) 0− 𝑠 : complex frequency variable, 𝑠 = 𝜎 + 𝑗𝜔 with 𝑠, 𝜔 are real numbers, 𝑠 ∈ 𝐶 for which makes 𝐹 𝑠 convergent ℒ : Laplace transform 𝐹(𝑠): a complex-valued function of complex numbers - The inverse Laplace transform of the function 𝐹(𝑠) for 𝑡 > is defined by the following relationship 𝜎+𝑗∞ 𝑓 𝑡 = ℒ −1 𝐹 𝑠 = 𝐹 𝑠 𝑒 𝑠𝑡 𝑑𝑠 = 𝑓 𝑡 𝑢(𝑡) (2.2) 2𝜋𝑗 𝜎−𝑗∞ 𝑢(𝑡) : the unit step function, 𝑢 𝑡 = HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑖𝑓 𝑡 > 0 𝑖𝑓 𝑡 < Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.07 Modeling in Frequency Domain §2.Laplace Transform Review - Ex.2.1 Laplace Transform of a Time Function Find the Laplace transform of 𝑓 𝑡 = 𝐴𝑒 −𝑎𝑡 𝑢(𝑡) Solution System Dynamics and Control 2.08 Modeling in Frequency Domain §2.Laplace Transform Review ∞ 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 𝐹 𝑠 = ∞ 𝐴𝑒 −𝑎𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = ∞ 𝑒 −(𝑎+𝑠)𝑡 𝑑𝑡 =𝐴 𝐴 𝑒 −(𝑎+𝑠)𝑡 𝑠+𝑎 𝐴 ⟹𝐹 𝑠 = 𝑠+𝑎 ∞ =− (2.3) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.09 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review - Ex.2.2 Inverse Laplace Transform Find the inverse Laplace transform of 𝐹1 𝑠 = 1/(𝑠 + 3)2 Solution 𝑓 𝑡 = ℒ −1 = 𝑡𝑢(𝑡) 𝑠 𝑓1 𝑡 = ℒ −1 (𝑠 + 3)2 = 𝑒 −3𝑡 𝑓 𝑡 ⟹ 𝑓1 𝑡 = 𝑒 −3𝑡 𝑡𝑢(𝑡) (Table 2.1 – 3) (Table 2.2 – 4) ℒ 𝑡𝑢 𝑡 = 𝑠 ℒ 𝑒 −𝑎𝑡 𝑓(𝑡) = 𝐹(𝑠 + 𝑎) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.11 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review Case Roots of the Denominator of 𝐹(𝑠) are Real and Distinct 𝐾1 𝐾2 𝐹 𝑠 = = + (2.8) (𝑠 + 1)(𝑠 + 2) 𝑠 + 𝑠 + lim [(2.8) × (𝑠 + 1)] 𝑠→−1 ⟹ lim 𝑠→−1 𝑠+2 = lim 𝐾1 + 𝑠→−1 (𝑠 + 1)𝐾2 𝑠+2 ⟹ 𝐾1 = lim [(2.8) × (𝑠 + 2)] 𝑠→−2 (𝑠 + 2)𝐾1 = lim + 𝐾2 𝑠→−2 𝑠+1 𝑠+1 ⟹ 𝐾2 = −2 2 ⟹𝐹 𝑠 = − ⟹ 𝑓 𝑡 = 2𝑒 −𝑡 − 2𝑒 −2𝑡 𝑢(𝑡) 𝑠+1 𝑠+2 ⟹ lim 𝑠→−2 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.10 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review Partial-Fraction Expansion 𝑠 + 2𝑠 + 6𝑠 + 𝐹1 𝑠 = 𝑠2 + 𝑠 + = (𝑠 + 1) + 𝑠 +𝑠+5 𝑑𝛿(𝑡) ⟹ 𝑓1 𝑡 = + 𝛿 𝑡 + ℒ −1 𝑑𝑡 𝑠 +𝑠+5 𝐹(𝑠) Using partial-fraction expansion to expand function like 𝐹(𝑠) into a sum of terms and then find the inverse Laplace transform for each term ℒ 𝛿 𝑡 =1 𝑑𝑓(𝑡) ℒ 𝑑𝑡 = 𝑠𝐹(𝑠) → ℒ 𝑑𝛿(𝑡) 𝑑𝑡 (Table 2.1 – 1) (Table 2.2 – 7) =𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.12 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review In general, given an 𝐹(𝑠) whose denominator has real and distinct roots, a partial-fraction expansion 𝑁(𝑠) 𝐹 𝑠 = 𝐷(𝑠) 𝑁(𝑠) = 𝑠 + 𝑝1 𝑠 + 𝑝2 … 𝑠 + 𝑝𝑖 … (𝑠 + 𝑝𝑛 ) 𝐾1 𝐾2 𝐾𝑖 𝐾𝑛 = + + ⋯+ +⋯+ (2.11) 𝑠 + 𝑝1 𝑠 + 𝑝2 𝑠 + 𝑝𝑖 𝑠 + 𝑝𝑛 To find 𝐾𝑖 • multiply (2.11) by 𝑠 + 𝑝𝑖 • let 𝑠 approach −𝑝𝑖 (2.10) Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.13 Modeling in Frequency Domain §2.Laplace Transform Review - Ex.2.3 Laplace Transform Solution of a Differential Equation Given the following differential equation, solve for 𝑦(𝑡) if all initial conditions are zero Use the Laplace transform 𝑑2𝑦 𝑑𝑦 + 12 + 32𝑦 = 32𝑢(𝑡) 𝑑𝑡 𝑑𝑡 Solution 32 𝑠 𝑌 𝑠 + 12𝑠𝑌 𝑠 + 32𝑌 𝑠 = 𝑠 32 32 ⟹𝑌 𝑠 = = 𝑠 𝑠 + 12𝑠 + 32 𝑠(𝑠 + 4)(𝑠 + 8) 𝐾1 𝐾2 𝐾3 = + + 𝑠 𝑠+4 𝑠+8 ℒ ℒ 𝑑𝑓 𝑑𝑡 𝑑2 𝑓 𝑑𝑡 (Table 2.2 – 7) = 𝑠𝐹 𝑠 − 𝑓(0− ) = 𝑠 𝐹 𝑠 − 𝑠𝑓(0− ) − 𝑓′(0− ) Nguyen Tan Tien 2.15 Modeling in Frequency Domain §2.Laplace Transform Review 32 𝑠 𝑠 + 12𝑠 + 32 = − + 𝑠 𝑠+4 𝑠+8 𝑦 𝑡 = − 2𝑒 −4𝑡 + 𝑒 −8𝑡 𝑢(𝑡) (2.20) The 𝑢(𝑡) in (2.20) shows that the response is zero until 𝑡 = Unless otherwise specified, all inputs to systems in the text will not start until 𝑡 = Thus, output responses will also be zero until 𝑡 = For convenience, the 𝑢(𝑡) notation will be eliminated from now on Accordingly, the output response 𝑦 𝑡 = − 2𝑒 −4𝑡 + 𝑒 −8𝑡 (2.21) 𝑌 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.17 Modeling in Frequency Domain §2.Laplace Transform Review 𝑌 𝑠 = Matlab Result [r,p,k] = residue([32],[1,12,32,0]) r = [1, −2, 1], p = [−8, −4, 0], k = [ ] 𝑘 𝑟1 𝐾2 = 𝑝2 𝑝3 = − + 𝑠+8 𝑠+4 𝑠 = −2 𝑠→−4 32 =1 𝑠(𝑠 + 4) 𝑠→−8 ⟹𝑌 𝑠 = − + 𝑠 𝑠+4 𝑠+8 𝐾2 = Hence 𝑦 𝑡 = − 2𝑒 −4𝑡 + 𝑒 −8𝑡 𝑢(𝑡) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.16 Modeling in Frequency Domain §2.Laplace Transform Review Run ch2p1 through ch2p8 in Appendix B Learn how to use MATLAB to • represent polynomials • find roots of polynomials • multiply polynomials, and • find partial-fraction expansions HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control ⟹ lim 𝑠→−1 Nguyen Tan Tien 2.18 Modeling in Frequency Domain 𝑠+2 = lim 𝐾1 + 𝑠→−1 (𝑠 + 1)𝐾2 𝑠+2 ⟹ 𝐾1 = lim [(2.23) × (𝑠 + 2)] 𝑠→−2 𝑠+1 ⟹ 𝐾2 = −2 ⟹ lim 𝑠→−2 HCM City Univ of Technology, Faculty of Mechanical Engineering 32 𝑠(𝑠 + 8) 𝑠→−1 1 + (−2) +1 𝑠 − (−8) 𝑠 − (−4) 𝑟 𝑠 − ( ) 𝑟 𝑝1 Modeling in Frequency Domain §2.Laplace Transform Review Case Roots of the Denominator of 𝐹(𝑠) are Real and Repeated 𝐹 𝑠 = (2.22) (𝑠 + 1)(𝑠 + 2)2 𝐾1 𝐾2 𝐾3 = + + (2.23) 𝑠 + (𝑠 + 2)2 𝑠 + lim [(2.23) × (𝑠 + 1)] 32 = − + 𝑠 + 12𝑠 + 32𝑠 𝑠 𝑠 + 𝑠 + 𝑌 𝑠 =0+1 2.14 §2.Laplace Transform Review 32 𝐾1 𝐾2 𝐾3 𝑌 𝑠 = = + + 𝑠(𝑠 + 4)(𝑠 + 8) 𝑠 𝑠+4 𝑠+8 Evaluate the residue 𝐾𝑖 32 𝐾1 = =1 (𝑠 + 4)(𝑠 + 8) 𝑠→0 (Table 2.2 – 8) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control Nguyen Tan Tien = lim 𝑠→−2 HCM City Univ of Technology, Faculty of Mechanical Engineering (𝑠 + 2)𝐾1 + 𝐾2 𝑠+1 Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.19 Modeling in Frequency Domain §2.Laplace Transform Review 𝐾1 𝐾2 𝐾3 𝐹 𝑠 = + + 𝑠 + (𝑠 + 2)2 𝑠 + 𝐾1 = 2, 𝐾2 = −2 (2.23) × (𝑠 + 2)2 (𝑠 + 2)2 𝐾1 ⟹ = + 𝐾2 + (𝑠 + 2)𝐾3 𝑠+1 𝑠+1 Differentiate (2.24) with respect to 𝑠 −2 (𝑠 + 2)𝐾1 = + 𝐾3 𝑠+1 𝑠+1 ⟹ 𝐾3 = −2 2 ⟹𝑌 𝑠 = − − 𝑠+1 𝑠+2 𝑠+2 Hence 𝑦 𝑡 = 2𝑒 −𝑡 − 2𝑡𝑒 −2𝑡 − 2𝑒 −2𝑡 (2.23) 𝐹 𝑠 = Matlab Result Modeling in Frequency Domain Matlab Result 2 = − 𝑠 + 5𝑠 + 8𝑠 + 𝑠 + 𝑠+2 − 𝑠+2 [r,p,k] = residue([2],[1,5,8,4]) r = [−2, −2, 2], p = [−2, −2, −1], k = [ ] 1 𝐹 𝑠 = + (−2) + (−2) +2 [𝑠 − (−2)]2 𝑠 − (−2) 𝑟 𝑠 − (−1) 𝑘 𝑟 𝑟1 =− 𝑠+2 𝑝1 𝑝2 (𝑠 + 1)(𝑠 + 2)2 (2.22) F=zpk([], [-1 -2 -2],2) F= -(s+1) (s+2)^2 Continuous-time zero/pole/gain model (2.26) §2.Laplace Transform Review 𝐹 𝑠 = Modeling in Frequency Domain (2.24) Nguyen Tan Tien 2.21 2.20 §2.Laplace Transform Review HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control 𝑝3 2 2−𝑠+2+𝑠+1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.22 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review In general, given an 𝐹(𝑠) whose denominator has real and distinct roots, a partial-fraction expansion 𝑁(𝑠) 𝐹 𝑠 = 𝐷(𝑠) 𝑁(𝑠) = 𝑠 + 𝑝1 𝑟 𝑠 + 𝑝2 … 𝑠 + 𝑝𝑖 … (𝑠 + 𝑝𝑛 ) 𝐾1 𝐾1 𝐾2 = + + ⋯+ 𝑠 + 𝑝1 𝑟 𝑠 + 𝑝1 𝑟−1 𝑠 + 𝑝1 𝐾2 𝐾𝑖 𝐾𝑛 + + ⋯+ + ⋯+ (2.27) 𝑠 + 𝑝2 𝑠 + 𝑝𝑖 𝑠 + 𝑝𝑛 To find 𝐾𝑖 𝑟 𝑟 • multiply (2.27) by 𝑠 + 𝑝1 to get 𝐹1 𝑠 = 𝑠 + 𝑝1 𝐹(𝑠) • let 𝑠 approach −𝑝𝑖 𝑑 𝑖−1 𝐹1 (𝑠) 𝐾𝑖 = 𝑖 = 1,2, … , 𝑟; 0! = 𝑖 − ! 𝑑𝑠 𝑖−1 𝑠→𝑝1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.23 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review Case Roots of the Denominator of 𝐹(𝑠) are Complex or Imaginary 𝐹 𝑠 = (2.30) 𝑠(𝑠 + 2𝑠 + 5) 𝐾1 𝐾2 𝑠 + 𝐾3 = + (2.31) 𝑠 𝑠 + 2𝑠 + lim[(2.31) × 𝑠] ⟹ 𝐾1 = 3/5 𝑠→0 First multiplying (2.31) by the lowest common denominator, 𝑠(𝑠 + 2𝑠 + 5), and clearing the fraction = 𝐾1 𝑠 + 2𝑠 + + 𝐾2 𝑠 + 𝐾3 𝑠 (2.32) ⟹ = 𝐾2 + 𝑠 + 𝐾3 + 𝑠 + (2.33) 5 Balancing the coefficients: 𝐾2 = −3/5, 𝐾3 = −6/5 31 𝑠 +2 𝐹 𝑠 = = − 𝑠(𝑠 + 2𝑠 + 5) 𝑠 𝑠 + 2𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.24 §2.Laplace Transform Review 31 𝑠 + 𝐹 𝑠 = − 𝑠 𝑠 + 2𝑠 + 3 𝑠 + + 1/2 = − 𝑠 (𝑠 + 1)2 +22 3 ⟹ 𝑓 𝑡 = − 𝑒 −𝑡 𝑐𝑜𝑠2𝑡 + 𝑠𝑖𝑛2𝑡 5 Nguyen Tan Tien Modeling in Frequency Domain (2.38) or 𝑓 𝑡 = 0.6 − 0.671𝑒 −𝑡 cos(2𝑡 − 𝜙) ℒ 𝐴𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 + 𝐵𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 = 𝐵 𝑠+𝑎 +𝐵𝜔 (𝑠+𝑎)2 +𝜔 HCM City Univ of Technology, Faculty of Mechanical Engineering (2.41) (Table 2.1 – 6&7) Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.25 Modeling in Frequency Domain §2.Laplace Transform Review (𝑠 + 1)(𝑠 + 2)2 𝑦 𝑡 = 2𝑒 −𝑡 − 2𝑡𝑒 −2𝑡 − 2𝑒 −2𝑡 Result 2.26 Modeling in Frequency Domain §2.Laplace Transform Review 𝐹 𝑠 = Matlab System Dynamics and Control 𝑝1 𝑠(𝑠 + 2𝑠 + 5) (2.30) (2.26) numf=2; denf=poly([-1 -2 -2]); [r,p,k]=residue(numf,denf) r = [-2 -2 2], p = [-2 -2 -1], k = [] 1 𝐹 𝑠 = + −2 + (−2) +2 [𝑠 − (−2)]2 𝑠 − (−2) 𝑟 𝑠 − (−1) 𝑘 𝑟 𝑟 𝐹 𝑠 = (2.22) 𝑝2 Matlab Result F=tf([3],[1 0]) F= s^3 + s^2 + s Continuous-time transfer function 𝑝3 1 = −2 −2 +2 (𝑠 + 2)2 𝑠+2 𝑠+1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.27 Modeling in Frequency Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.28 §2.Laplace Transform Review In general, given an 𝐹(𝑠) whose denominator has complex or purely imaginary roots, a partial-fraction expansion 𝑁(𝑠) 𝐹 𝑠 = 𝐷(𝑠) 𝑁(𝑠) = 𝑠 + 𝑝1 (𝑠 + 𝑎𝑠 + 𝑏) … 𝐾1 𝐾2 𝑠 + 𝐾3 = + +⋯ (2.42) (𝑠 + 𝑝1 ) (𝑠 + 𝑎𝑠 + 𝑏) To find 𝐾𝑖 • the 𝐾𝑖 ’s in (2.42) are found through balancing the coefficients of the equation after clearing fractions • put (𝐾2 𝑠 + 𝐾3 )/(𝑠 + 𝑎𝑠 + 𝑏) in to the form 𝐵 𝑠 + 𝑎 + 𝐵𝜔 (𝑠 + 𝑎)2 +𝜔 §2.Laplace Transform Review HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.29 Modeling in Frequency Domain §2.Laplace Transform Review (2.30) 𝑠(𝑠 + 2𝑠 + 5) 3/5 + 𝑗1 − 𝑗1 (2.47) 𝐹 𝑠 = − + 𝑠 20 𝑠 + + 𝑗2 𝑠 + + 𝑗2 𝐹 𝑠 = Matlab Result numf=3; denf=[1 0]; [r,p,k]=residue(numf,denf) r=[-0.3+0.15i; -0.3-0.15i; 0.6]; p=[-1+2i; -1-2i; 0]; k=[] 𝐹 𝑠 = + (−0.3 + 𝑗0.15) 𝑠 − (−1 + 𝑗2) 𝑘 𝑟1 Nguyen Tan Tien Modeling in Frequency Domain 𝑠(𝑠 + 2𝑠 + 5) 3 𝑓 𝑡 = − 𝑒 −𝑡 𝑐𝑜𝑠2𝑡 + 𝑠𝑖𝑛2𝑡 5 𝐹 𝑠 = Matlab Result (2.30) (2.38) syms s; f=ilaplace(3/(s*(s^2+2*s+5))); pretty(f) f= 3/5 - (3*exp(-t)*(cos(2*t) + sin(2*t)/2))/5 / sin(2 t) \ exp(-t) | cos(2 t) + | 3 \ / - - -5 System Dynamics and Control 2.30 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review Run ch2sp1 and ch2sp2 in Appendix F Learn how to use the Symbolic Math Toolbox to • construct symbolic objects • find the inverse Laplace transforms of frequency functions • find the Laplace of time functions 𝑝1 1 +(−0.3 − 𝑗0.15) + (0.6) 𝑠 − (−1 − 2𝑗) 𝑠 − (0) 𝑟 𝑟2 𝑝2 𝑝3 0.3 − 𝑗0.15 0.3 + 𝑗0.15 =− − + 0.6 𝑠 + − 2𝑗 𝑠 + + 2𝑗 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.31 Modeling in Frequency Domain §2.Laplace Transform Review Skill-Assessment Ex.2.1 Problem Find the Laplace transform of 𝑓 𝑡 = 𝑡𝑒 −5𝑡 Solution 𝐹 𝑠 = ℒ 𝑡𝑒 −5𝑡 = (𝑠 + 5)2 Matlab Result syms t s F; f = t*exp(-5*t); F=laplace(f, s); pretty(F) (s + 5) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.33 Nguyen Tan Tien Modeling in Frequency Domain §2.Laplace Transform Review 51 10 40 𝐹 𝑠 = −5 + + 9𝑠 𝑠 + (𝑠 + 3)2 𝑠 + Taking the inverse Laplace transform 10 40 𝑓 𝑡 = − 5𝑒 −2𝑡 + 𝑡𝑒 −3𝑡 + 𝑒 −3𝑡 9 Matlab Result syms s; f=ilaplace(10/(s*(s+2)*(s+3)^2)); pretty(f) exp(-3 t) 40 t exp(-3 t) 10 - - exp(-2 t) + + 9 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.35 Nguyen Tan Tien Modeling in Frequency Domain System Dynamics and Control 2.32 Modeling in Frequency Domain §2.Laplace Transform Review Skill-Assessment Ex.2.2 Problem Find the inverse Laplace transform of 10 𝐹 𝑠 = 𝑠(𝑠 + 2)(𝑠 + 3)2 Solution Expanding 𝐹(𝑠) by partial fractions 𝐴 𝐵 𝐶 𝐷 𝐹 𝑠 = + + + 𝑠 𝑠 + (𝑠 + 3)2 𝑠 + where, 10 10 𝐴= = ,𝐵 = (𝑠 + 2)(𝑠 + 3)2 𝑠→0 𝑠(𝑠 + 3)2 = −5 𝑠→−2 10 10 𝑑𝐹(𝑠) 40 = , 𝐷 = (𝑠 + 3)2 = 𝑠(𝑠 + 2) 𝑠→−3 𝑑𝑠 𝑠→−3 51 10 40 ⟹𝐹 𝑠 = −5 + + 9𝑠 𝑠 + (𝑠 + 3)2 𝑠 + 𝐶= HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.34 Nguyen Tan Tien Modeling in Frequency Domain §3.The Transfer Function - The transfer function of a component is the quotient of the Laplace transform of the output divided by the Laplace transform of the input, with all initial conditions assumed to be zero - Transfer functions are defined only for linear time invariant systems - The input-output relationship of a control system 𝐺 𝑠 𝐶(𝑠) 𝑑 𝑛 𝑐(𝑡) 𝑑 𝑛−1 𝑐 𝑡 𝑎𝑛 + 𝑎𝑛−1 + ⋯ + 𝑎0 𝑐 𝑡 𝑛 𝑑𝑡 𝑑𝑡 𝑛−1 𝑚 𝑑 𝑟(𝑡) 𝑑 𝑚−1 𝑟(𝑡) = 𝑏𝑚 + 𝑏𝑚−1 + ⋯ + 𝑏0 𝑟 𝑡 𝑑𝑡 𝑚 𝑑𝑡 𝑚−1 𝑐(𝑡): output 𝑟(𝑡): input 𝑎𝑖 ’s, 𝑏𝑖 ’s : constant HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.36 Nguyen Tan Tien Modeling in Frequency Domain §3.The Transfer Function 𝑑 𝑛 𝑐(𝑡) 𝑑 𝑛−1 𝑐 𝑡 𝑎𝑛 + 𝑎𝑛−1 + ⋯ + 𝑎0 𝑐 𝑡 𝑛 𝑑𝑡 𝑑𝑡 𝑛−1 𝑚 𝑑 𝑟(𝑡) 𝑑 𝑚−1 𝑟(𝑡) = 𝑏𝑚 + 𝑏𝑚−1 + ⋯ + 𝑏0 𝑟 𝑡 𝑚 𝑑𝑡 𝑑𝑡 𝑚−1 - Taking the Laplace transform of both sides with zero initial conditions 𝑎𝑛 𝑠 𝑛 𝐶 𝑠 + 𝑎𝑛−1 𝑠 𝑛−1 𝐶 𝑠 + ⋯ + 𝑎0 𝐶 𝑠 = 𝑏𝑚 𝑠 𝑚 𝑅 𝑠 + 𝑏𝑚−1 𝑠 𝑚−1 𝑅 𝑠 + ⋯ + 𝑏0 𝑅 𝑠 - The transfer function 𝐶(𝑠) 𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 + ⋯ + 𝑏0 𝐺 𝑠 = = (2.53) 𝑅(𝑠) 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + ⋯ + 𝑎0 - The output of the system can be written in the form 𝐶 𝑠 = 𝐺 𝑠 𝑅(𝑠) (2.54) §3.The Transfer Function - Ex.2.4 Transfer Function for a Differential Equation Find the transfer function represented by 𝑑𝑐(𝑡) + 2𝑐(𝑡) = 𝑟(𝑡) 𝑑𝑡 Solution Taking the Laplace transform with zero initial conditions 𝑠𝐶 𝑠 + 2𝐶(𝑠) = 𝑅(𝑠) The transfer function 𝐶(𝑠) 𝐺 𝑠 = = 𝑅(𝑠) 𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.37 Modeling in Frequency Domain System Dynamics and Control 2.38 Modeling in Frequency Domain §3.The Transfer Function Run ch2p9 through ch2p12 in Appendix B Learn how to use MATLAB to • create transfer functions with numerators and denominators in polynomial or factored form • convert between polynomial and factored forms ã plot time functions Đ3.The Transfer Function Run ch2sp3 in Appendix F Learn how to use the Symbolic Math Toolbox to • simplify the input of complicated transfer functions as well as improve readability • enter a symbolic transfer function and convert it to a linear time-invariant (LTI) object as presented in Appendix B, ch2p9 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.39 Nguyen Tan Tien Modeling in Frequency Domain §3.The Transfer Function - Ex.2.5 System Response from the Transfer Function Given 𝐺 𝑠 = 1/(𝑠 + 2), find the response, 𝑐(𝑡) to an input, 𝑟 𝑡 = 𝑢(𝑡), a unit step, assuming zero initial conditions Solution For a unit step 𝑟 𝑡 = 𝑢 𝑡 ⟹ 𝑅 𝑠 = 1/𝑠 The output 1 𝐶 𝑠 =𝑅 𝑠 𝐺 𝑠 = 𝑠𝑠+2 Expanding by partial fractions 11 1 𝐶 𝑡 = − 2𝑠 2𝑠 + Taking the inverse Laplace transform 𝑐 𝑡 = 0.5 − 0.5𝑒 −2𝑡 2.41 Nguyen Tan Tien Modeling in Frequency Domain §3.The Transfer Function 𝑐 𝑡 = Matlab 2.40 Modeling in Frequency Domain §3.The Transfer Function 𝑠+2 𝑅 𝑠 = 𝑠 1 𝑐 𝑡 = − 𝑒 −2𝑡 2 𝐺 𝑠 = Matlab Result (2.60) syms s C=1/(s*(s+2)) C=ilaplace(C) C= 1/2 - exp(-2*t)/2 𝑐 𝑡 = 1 −2𝑡 − 𝑒 2 (2.60) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control Nguyen Tan Tien 1 −2𝑡 − 𝑒 2 (2.60) t=0:0.01:1; plot(t,(1/2-1/2*exp(-2*t))) Result HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.42 Nguyen Tan Tien Modeling in Frequency Domain §3.The Transfer Function Skill-Assessment Ex.2.3 Problem Find the transfer function, 𝐺 𝑠 = 𝐶(𝑠)/𝑅(𝑠), corresponding to the differential equation 𝑑3𝑐 𝑑2𝑐 𝑑𝑐 𝑑2𝑟 𝑑𝑟 + + + 5𝑐 = + + 3𝑟 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 Solution Taking the Laplace transform with zero initial conditions 𝑠 𝐶 𝑠 + 3𝑠 𝐶 𝑠 + 7𝑠𝐶 𝑠 + 5𝐶 𝑠 = 𝑠 𝑅 𝑠 + 4𝑠𝑅 𝑠 + 3𝑅 𝑠 Collecting terms 𝑠 + 3𝑠 + 7𝑠 + 𝐶 𝑠 = 𝑠 + 4𝑠 + 𝑅(𝑠) The transfer function 𝐶(𝑠) 𝑠2 + 4𝑠 + 𝐺 𝑠 = = 𝑅(𝑠) 𝑠3 + 3𝑠2 + 7𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.43 Modeling in Frequency Domain System Dynamics and Control 2.44 Modeling in Frequency Domain §3.The Transfer Function Skill-Assessment Ex.2.4 Problem Find the differential equation corresponding to the transfer function 2𝑠 + 𝐺 𝑠 = 𝑠 + 6𝑠 + Solution The transfer function 𝐶(𝑠) 2𝑠 + 𝐺 𝑠 = = 𝑅(𝑠) 𝑠 + 6𝑠 + Cross multiplying 𝑑2𝑐 𝑑𝑐 𝑑𝑟 + + 2𝑐 = + 𝑟 𝑑𝑡 𝑑𝑡 𝑑𝑡 §3.The Transfer Function Skill-Assessment Ex.2.5 Problem Find the ramp response for a system whose transfer function 𝑠 𝐺 𝑠 = (𝑠 + 4)(𝑠 + 8) Solution For a ramp response 𝑟 𝑡 = 𝑡𝑢 𝑡 ⟹ 𝑅 𝑠 = 𝑠 The output 𝐶 𝑠 =𝑅 𝑠 𝐺 𝑠 𝑠 = 𝑠 (𝑠 + 4)(𝑠 + 8) 𝐴 𝐵 𝐶 = + + 𝑠 𝑠+4 𝑠+8 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.45 Modeling in Frequency Domain §3.The Transfer Function 𝐴 𝐵 𝐶 𝐶 𝑠 = = + + 𝑠(𝑠 + 4)(𝑠 + 8) 𝑠 𝑠 + 𝑠 + where, 1 𝐴= = (𝑠 + 4)(𝑠 + 8) 𝑠→0 32 𝐵= 𝑠(𝑠 + 8) =− 𝑠→−4 System Dynamics and Control 2.46 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Summarizes the components and the relationships between voltage and current and between voltage and charge under zero initial conditions 16 1 = 𝑠(𝑠 + 4) 𝑠→−8 32 1 1 1 ⟹𝐶 𝑠 = − + 32 𝑠 16 𝑠 + 32 𝑠 + The ramp response 1 𝑐 𝑡 = − 𝑒 −4𝑡 + 𝑒 −8𝑡 32 16 32 𝐶= HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.47 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Simple Circuits via Mesh Analysis - Ex.2.6 Transfer Function - Single Loop via the Differential Equation Find the transfer function 𝑉𝐶 (𝑠)/𝑉(𝑠) Solution The voltage loop 𝑑𝑖 1 𝐿 + 𝑅𝑖 + 𝑖 𝜏 𝑑𝜏 = 𝑣(𝑡) 𝑑𝑡 𝐶 Using the relationships 𝑖 𝑡 = 𝑑𝑞(𝑡)/𝑑𝑡 and 𝑞 = 𝐶𝑣𝐶 𝑑2𝑞 𝑑𝑞 𝐿 +𝑅 + 𝑞 = 𝑣(𝑡) 𝑑𝑡 𝑑𝑡 𝐶 𝑑 𝑣𝐶 𝑑𝑣𝐶 ⟹ 𝐿𝐶 + 𝑅𝐶 + 𝑣𝐶 = 𝑣(𝑡) 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.48 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions 𝑑 𝑣𝐶 𝑑𝑣𝐶 𝐿𝐶 + 𝑅𝐶 + 𝑣𝐶 = 𝑣(𝑡) 𝑑𝑡 𝑑𝑡 Taking Laplace transform assuming zero initial conditions 𝐿𝐶𝑠 + 𝑅𝐶𝑠 + 𝑉𝐶 𝑠 = 𝑉(𝑠) Solving for the transfer function 𝑉𝐶 (𝑠) 𝐿𝐶 = 𝑅 𝑉(𝑠) 𝑠 + 𝑠 + 𝐿 𝐿𝐶 Block diagram of series RLC electrical network HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.49 Modeling in Frequency Domain §4.Electrical Network Transfer Functions Impedance - A resistance resists or “impedes” the flow of current The corresponding relation is 𝑣/𝑖 = 𝑅 Capacitance and inductance elements also impede the flow of current - In electrical systems an impedance is a generalization of the resistance concept and is defined as the ratio of a voltage transform 𝑉(𝑠) to a current transform 𝐼(𝑠) and thus implies a current source - Standard symbol for impedance 𝑉(𝑠) 𝑍(𝑠) ≡ (2.70) 𝐼(𝑠) - Kirchhoff’s voltage law to the transformed circuit [ Sum of Impedances ] × 𝐼 𝑠 = [ Sum of Applied Voltages ] (2.72) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.51 Nguyen Tan Tien Modeling in Frequency Domain System Dynamics and Control 2.50 Modeling in Frequency Domain §4.Electrical Network Transfer Functions - The impedance of a resistor is its resistance 𝑍 𝑠 =𝑅 - For a capacitor 𝑡 𝐼(𝑠) 𝑣 𝑡 = 𝑖𝑑𝑡 ⟹ 𝑉 𝑠 = 𝐶 𝐶 𝑠 𝑠 The impedance of a capacitor 𝑍 𝑠 = 𝐶𝑠 - For an inductor 𝑑𝑖 𝑣 𝑡 = 𝐿 ⟹ 𝑉 𝑠 = 𝐿𝐼 𝑠 𝑠 𝑑𝑡 The impedance of a inductor 𝑍 𝑠 = 𝐿𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.52 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Series and Parallel Impedances - The concept of impedance is useful because the impedances of individual elements can be combined with series and parallel laws to find the impedance at any point in the system - The laws for combining series or parallel impedances are extensions to the dynamic case of the laws governing series and parallel resistance elements §4.Electrical Network Transfer Functions - Series Impedances • Two impedances are in series if they have the same current If so, the total impedance is the sum of the individual impedances 𝑖 𝑣 𝑅 𝐶 𝑍 𝑠 = 𝑍1 𝑠 + 𝑍2 (𝑠) • Example: a resistor 𝑅 and capacitor 𝐶 in series have the equivalent impedance 𝑅𝐶𝑠 + 𝑍 𝑠 =𝑅+ = 𝐶𝑠 𝐶𝑠 𝑉(𝑠) 𝑅𝐶𝑠 + ⟹ ≡𝑍 𝑠 = 𝐼(𝑠) 𝐶𝑠 and the differential equation model is 𝑑𝑣 𝑑𝑖 𝐶 = 𝑅𝐶 + 𝑖(𝑡) 𝑑𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.53 Nguyen Tan Tien Modeling in Frequency Domain System Dynamics and Control 2.54 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions - Parallel Impedances • Two impedances are in parallel if they have the same voltage difference across them Their impedances combine by the reciprocal rule 1 = + 𝑍(𝑠) 𝑍1 (𝑠) 𝑍2 (𝑠) • Example: a resistor 𝑅 and capacitor 𝐶 in parallel have the equivalent impedance 1 𝑉(𝑠) 𝑅 = + ⟹ ≡𝑍 𝑠 = 𝑍(𝑠) 1/𝐶𝑠 𝑅 𝐼(𝑠) 𝑅𝐶𝑠 + and the differential equation model is 𝑑𝑣 𝑅𝐶 + 𝑣 = 𝑅𝑖(𝑡) 𝑑𝑡 §4.Electrical Network Transfer Functions Admittance 𝐼(𝑠) 𝑌 𝑠 ≡ = 𝑍(𝑠) 𝑉(𝑠) In general, admittance is complex • The real part of admittance is called conductance 𝐺= 𝑅 • The imaginary part of admittance is called susceptance When we take the reciprocal of resistance to obtain the admittance, a purely real quantity results The reciprocal of resistance is called conductance HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 1/11/2016 System Dynamics and Control 2.55 Modeling in Frequency Domain System Dynamics and Control 2.56 Modeling in Frequency Domain §4.Electrical Network Transfer Functions Instead of taking the Laplace transform of the differential equation, we can draw the transformed circuit and obtain the Laplace transform of the differential equation simply by applying Kirchhoff’s voltage law to the transformed circuit The steps are as follows 1.Redraw the original network showing all time variables, such as 𝑣(𝑡), 𝑖(𝑡), and 𝑣𝐶 (𝑡), as Laplace transforms 𝑉(𝑠), 𝐼(𝑠), and 𝑉𝐶 (𝑠), respectively 2.Replace the component values with their impedance values This replacement is similar to the case of dc circuits, where we represent resistors with their resistance values §4.Electrical Network Transfer Functions - Ex.2.7 Transfer Function - Single Loop via Transform Method Find the transfer function 𝑉𝐶 (𝑠)/𝑉(𝑠) Solution The mess equation using impedances 𝐿𝑠 + 𝑅 + 𝐼 𝑠 = 𝑉(𝑠) 𝐶𝑠 𝐼(𝑠) ⟹ = 𝑉(𝑠) 𝐿𝑠 + 𝑅 + 𝐶𝑠 The voltage across the capacitor 𝑉𝐶 𝑠 = 𝐼(𝑠) 𝐶𝑠 ⟹ 𝑉𝐶 (𝑠)/𝑉(𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.57 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Simple Circuits via Nodal Analysis - Ex.2.8 Transfer Function - Single Node via Transform Method Find the transfer function 𝑉𝐶 (𝑠)/𝑉(𝑠) Solution The transfer function can be obtained by summing currents flowing out of the node whose voltage is 𝑉𝐶 (𝑠) 𝑉𝐶 (𝑠) 𝑉𝐶 (𝑠) 𝑉𝐶 𝑠 − 𝑉(𝑠) + =0 ⟹ 𝑉(𝑠) 1/𝐶𝑠 𝑅 + 𝐿𝑠 𝑉𝐶 (𝑠) : the current flowing out of the node through the 𝐼/𝐶𝑠 capacitor System Dynamics and Control 2.58 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Complex Circuits via Mesh Analysis To solve complex electrical networks - those with multiple loops and nodes – using mesh analysis 1.Replace passive element values with their impedances 2.Replace all sources and time variables with their Laplace transform 3.Assume a transform current and a current direction in each mesh 4.Write Kirchhoff’s voltage law around each mesh 5.Solve the simultaneous equations for the output 6.Form the transfer function 𝑉𝐶 𝑠 − 𝑉(𝑠) : the current flowing out of the node through the 𝑅 + 𝐿𝑠 series resistor and inductor HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.59 Nguyen Tan Tien Modeling in Frequency Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.60 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions - Ex.2.10 Transfer Function – Multiple Loops Find the transfer function 𝐼2 (𝑠)/𝑉(𝑠) Solution Convert the network into Laplace transforms Summing voltages around each mesh through which the assumed currents flow 𝑅1 𝐼1 + 𝐿𝑠𝐼1 − 𝐿𝑠𝐼2 = 𝑉 𝐿𝑠𝐼2 + 𝑅2𝐼2 + 𝐼2 − 𝐿𝑠𝐼1 = 𝐶𝑠 or 𝑅1 + 𝐿𝑠 𝐼1 − 𝐿𝑠𝐼2 = 𝑉 (2.80) −𝐿𝑠𝐼1 + 𝐿𝑠 + 𝑅2 + 𝐼 =0 𝐶𝑠 §4.Electrical Network Transfer Functions Cramer’s Rule Consider a system of 𝑛 linear equations for 𝑛 unknowns, represented in matrix multiplication form as follows 𝐴𝑥 = 𝑏 𝐴 : (𝑛 × 𝑛) matrix has a nonzero determinant 𝑥 : the column vector of the variables 𝑥 = (𝑥1 , … , 𝑥𝑛 )𝑇 𝑏 : the column vector of known parameters The system has a unique solution, whose individual values for the unknowns are given by det(𝐴𝑖 ) 𝑥𝑖 = , 𝑖 = 1, … , 𝑛 det(𝐴) 𝐴𝑖 : the matrix formed by replacing the 𝑖th column of 𝐴 by the column vector 𝑏 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 10 1/11/2016 System Dynamics and Control 2.79 Modeling in Frequency Domain §4.Electrical Network Transfer Functions Inverting Operational Amplifiers 𝑍𝑖 𝑠 = ∞ → 𝐼𝑎 (𝑠) = 𝐼1 𝑠 = −𝐼2 𝑠 𝐴=∞ → 𝑣1 𝑡 ≈ 𝑉𝑖 𝑠 𝑉𝑜 𝑠 𝐼1 𝑠 = = −𝐼2 𝑠 = − 𝑍1 𝑠 𝑍2 𝑠 The transfer function of the inverting operational amplifier 𝑉𝑜 (𝑠) 𝑍2 (𝑠) =− (2.97) 𝑉𝑖 (𝑠) 𝑍1 (𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.81 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Noninverting Operational Amplifiers 𝑉𝑜 = 𝐴 𝑉𝑖 − 𝑉1 𝑍1 𝑉1 = 𝑉 𝑍1 + 𝑍2 𝑜 The transfer function of the noninverting operational amplifier 𝑉𝑜 (𝑠) 𝑍1 𝑠 + 𝑍2 (𝑠) = (2.104) 𝑉𝑖 (𝑠) 𝑍1 (𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.83 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Skill Assessment Ex.2.6 Problem Find 𝐺 𝑠 = 𝑉𝐿 (𝑠)/𝑉(𝑠) using mesh and nodal analysis Solution Mesh analysis Writing the mesh equations 𝑠 + 𝐼1 − 𝑠𝐼2 − 𝐼3 = 𝑉 −𝑠𝐼1 + 2𝑠 + 𝐼2 − 𝐼3 = −𝐼1 − 𝐼2 + 𝑠 + 𝐼3 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 2.80 Modeling in Frequency Domain §4.Electrical Network Transfer Functions - Ex.2.14 Transfer Function – Inverting Op-Amp Circuit Find the transfer function 𝑉𝑜 (𝑠)/𝑉𝑖 (𝑠) Solution The impedances 1 1 360 × 103 = + → 𝑍1 = = 2.016𝑠 + 𝑍1 1/𝐶1𝑠 𝑅1 𝐶1𝑠 + 𝑅1 10 𝑍2 = 𝑅2 + = 220 × 10 + 𝐶2 𝑠 𝑠 The transfer function 360 × 103 𝑉𝑜(𝑠) 𝑍2 𝑠 𝑠2 + 45.95𝑠 + 22.55 =− = − 2.016𝑠 + = −1.232 10 𝑉𝑖(𝑠) 𝑍1 𝑠 𝑠 220 × 103 + 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.82 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions - Ex.2.15 Transfer Function – Noninverting Op-Amp Circuit Find the transfer function 𝑉𝑜 (𝑠)/𝑉𝑖 (𝑠) Solution The impedances 𝑍1 = 𝑅1 + 𝐶1𝑠 𝑅2 𝐶2𝑠 𝑍2 = 𝑅2 + 𝐶2𝑠 The transfer function 𝑉𝑜 (𝑠) 𝑍1 𝑠 + 𝑍2(𝑠) 𝐶2𝐶1𝑅2𝑅1𝑠2 + (𝐶2𝑅2 + 𝐶1𝑅2 + 𝐶1𝑅1)𝑠 + = = 𝑉𝑖 (𝑠) 𝑍1(𝑠) 𝐶2𝐶1𝑅2𝑅1𝑠2 + 𝐶2𝑅2 + 𝐶1𝑅1 𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.84 Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions 𝑠 + 𝐼1 − 𝑠𝐼2 − 𝐼3 = 𝑉 −𝑠𝐼1 + 2𝑠 + 𝐼2 − 𝐼3 = −𝐼1 − 𝐼2 + 𝑠 + 𝐼3 = Solving the mesh equation for 𝐼2 𝑠 + 𝑉 −1 −𝑠 −1 𝑠 + 2𝑠 + 𝑉 −1 𝑠 + 𝐼2 = = 𝑠+1 −𝑠 −1 𝑠(𝑠 + 5𝑠 + 2) −𝑠 2𝑠 + −1 −1 −1 𝑠+2 The voltage across 𝐿 𝑠 + 2𝑠 + 𝑉 𝑉𝐿 = 𝑠𝐼2 = 𝑠 + 5𝑠 + 𝑉𝐿 𝑠 + 2𝑠 + ⟹𝐺 𝑠 = = 𝑉 𝑠 + 5𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 14 1/11/2016 System Dynamics and Control 2.85 Modeling in Frequency Domain §4.Electrical Network Transfer Functions Nodal analysis Modeling in Frequency Domain Solving the nodal equation for 𝑉𝐿 +2 𝑉 𝑠 −1 𝑠 + 2𝑠 + 𝑉 𝑠 𝑉𝐿 = = 𝑠 + 5𝑠 + + −1 𝑠 −1 +1 𝑠 𝑉𝐿 𝑠 + 2𝑠 + ⟹𝐺 𝑠 = = 𝑉 𝑠 + 5𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering 2.87 2.86 §4.Electrical Network Transfer Functions + 𝑉1 − 𝑉𝐿 = 𝑉 𝑠 −𝑉1 + + 𝑉𝐿 = 𝑉 𝑠 𝑠 Writing the nodal equations + 𝑉1 − 𝑉𝐿 = 𝑉 𝑠 −𝑉1 + + 𝑉𝐿 = 𝑉 𝑠 𝑠 System Dynamics and Control System Dynamics and Control Nguyen Tan Tien Modeling in Frequency Domain §4.Electrical Network Transfer Functions Skill Assessment Ex.2.7 Problem If 𝑍1 (𝑠)is the impedance of a 10𝜇𝐹 capacitor and 𝑍2 (𝑠) is the impedance of a 100𝑘Ω resistor, find the transfer function, 𝐺 𝑠 = 𝑉𝑜 (𝑠)/𝑉𝑖 (𝑠) if these components are used with (a) an inverting op-amp and (b) a noninverting op-amp HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.88 Modeling in Frequency Domain §4.Electrical Network Transfer Functions (a) Inverting Op-Amp 𝑍2 𝑍1 105 =− 10 𝑠 = −𝑠 𝐺 𝑠 =− (b) Noninverting Op-Amp 𝐺 𝑠 = Solution 1 105 = −5 = 𝐶𝑠 10 𝑠 𝑠 𝑍2 = 𝑍𝑅 = 𝑅 = 105 = 105 𝑠 =𝑠+1 𝑍1 = 𝑍𝐶 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.89 Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑍1 + 𝑍2 𝑍1 105 + 105 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.90 Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions - Ex.2.16 Transfer Function - One Equation of Motion Find the transfer function 𝑋(𝑠)/𝐹(𝑠) Solution Free body diagram Using Newton’s law to sum all of the forces 𝑑2 𝑥(𝑡) 𝑑𝑥(𝑡) 𝑀 + 𝑓𝑣 + 𝐾𝑥 𝑡 = 𝑓(𝑡) 𝑑𝑡 𝑑𝑡 Taking Laplace transform 𝑀𝑠 𝑋 𝑠 + 𝑓𝑣 𝑠𝑋 𝑠 + 𝐾𝑋 𝑠 = 𝐹(𝑠) The transfer function 𝑋(𝑠) 𝐹 𝑠 𝐺(𝑠) = = 𝐹(𝑠) 𝑀𝑠 + 𝑓𝑣 𝑠 + 𝐾 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 15 1/11/2016 System Dynamics and Control 2.91 Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions Impedance - Define impedance for mechanical components 𝐹 𝑠 𝑍𝑀 𝑠 ≡ 𝑋 𝑠 ⟹ 𝐹 𝑠 = 𝑍𝑀 𝑠 𝑋 𝑠 System Dynamics and Control 2.92 Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions - Ex.2.17 Transfer Function - Two Degrees of Freedom Find the transfer function 𝑋2 (𝑠)/𝐹(𝑠) Solution Free body diagram of 𝑀1 Sum of Impedances × 𝑋(𝑠) = Sum of Applied Forces - The impedance of a spring is its stiffness coefficient 𝐹 𝑠 = 𝐾𝑋 𝑠 ⟹ 𝑍𝑀 𝑠 = 𝐾 - For the viscous damper 𝐹 𝑠 = 𝑓𝑣 𝑠𝑋 𝑠 ⟹ 𝑍𝑀 𝑠 = 𝑓𝑣 𝑠 - For the mass 𝐹 𝑠 = 𝑀𝑠 𝑋 𝑠 ⟹ 𝑍𝑀 𝑠 = 𝑀𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.93 (2.112) (2.113) (2.114) Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions The Laplace transform of the equation of motion of 𝑀1 + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋2 = 𝐹 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.94 Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions The Laplace transform of the equations of motion + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋2 = 𝐹 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋1 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + 𝐾2 + 𝐾3 𝑋2 = Free body diagram of 𝑀2 The Laplace transform of the equations of motion 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 𝑠 + 𝐾1 + 𝐾2 𝐹 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋2 = ∆ = 𝑓𝑣3 𝑠 + 𝐾2 𝐹 ∆ where ∆= The Laplace transform of the equation of motion of 𝑀2 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋1 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + 𝐾2 + 𝐾3 𝑋2 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.95 Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions Note The Laplace transform of the equations of motion of 𝑀1 + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋2 = 𝐹 The Laplace transform of the equations of motion of 𝑀2 − 𝑓𝑣3 𝑠 + 𝐾2 𝑋1 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + 𝐾2 + 𝐾3 𝑋2 = 𝑀1𝑠2 + 𝑓𝑣1 +𝑓𝑣3 𝑠+ 𝐾1 +𝐾2 − 𝑓𝑣3 𝑠 + 𝐾2 − 𝑓𝑣3 𝑠 + 𝐾2 𝑀2𝑠2 + 𝑓𝑣2 +𝑓𝑣3 𝑠+ 𝐾2 +𝐾3 The transfer function 𝑋2 (𝑠) 𝑓𝑣3 𝑠 + 𝐾2 𝐺 𝑠 = = 𝐹(𝑠) ∆ HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.96 Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions - Ex.2.18 Equations of Motion by Inspection Write the equations of motion for the mechanical network Solution The Laplace transform of the equations of motion of 𝑀1 + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 𝑠 + 𝐾1 + 𝐾2 𝑋1 − 𝐾2 𝑋2 − 𝑓𝑣3 𝑋3 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 16 1/11/2016 System Dynamics and Control 2.97 Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions −𝐾2 𝑋1 + [𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣4 𝑠 + 𝐾2 ]𝑋2 − 𝑓𝑣4 𝑠𝑋3 = 𝐹 System Dynamics and Control 2.99 2.98 Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions The Laplace transform of the equations of motion of 𝑀2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control The Laplace transform of the equations of motion of 𝑀3 −𝑓𝑣3 𝑠𝑋1 − 𝑓𝑣4 𝑠𝑋2 + [𝑀3 𝑠 + 𝑓𝑣3 + 𝑓𝑣4 𝑠]𝑋3 = Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions Skill-Assessment Ex.2.8 Problem Find the transfer function 𝑋2 (𝑠) 𝐺 𝑠 = 𝐹(𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.100 Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions Solution + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + 𝐾 𝑋1 − 𝑓𝑣1 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + 𝐾 𝑋1 − 𝑓𝑣1 + 𝑓𝑣2 + 𝑓𝑣3 𝑠 + 𝐾 𝑋2 = 𝐹 ⟹ +(𝑠 + 3𝑠 + 1)𝑋1 − (3𝑠 + 1)𝑋2 = 𝐹 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.101 Nguyen Tan Tien Modeling in Frequency Domain §5.Translational Mechanical System Transfer Functions + 𝑀2 𝑠 + 𝑓𝑣1 + 𝑓𝑣2 + 𝑓𝑣3 + 𝑓𝑣4 𝑠 + 𝐾 𝑋2 = ⟹ −(3𝑠 + 1)𝑋1 + (𝑠 + 4𝑠 + 1)𝑋2 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.102 Nguyen Tan Tien Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions + 𝑠 + 3𝑠 + 𝑋1 − 3𝑠 + 𝑋2 = 𝐹 −(3𝑠 + 1)𝑋1 + (𝑠 + 4𝑠 + 1)𝑋2 = The solution for 𝑋2 𝑠 + 3𝑠 + 𝐹 3𝑠 + 𝐹 − 3𝑠 + 𝑋2 = = ∆ ∆ where 𝑠 + 3𝑠 + − 3𝑠 + ∆= − 3𝑠 + 𝑠 + 4𝑠 + = 𝑠(𝑠 + 7𝑠 + 5𝑠 + 1) 𝑋2 (𝑠) 3𝑠 + ⟹𝐺 𝑠 = = 𝐹(𝑠) 𝑠(𝑠 + 7𝑠 + 5𝑠 + 1) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 17 1/11/2016 System Dynamics and Control 2.103 Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions - Ex.2.19 Transfer Function – Two Equations of Motion Find the transfer function, 𝜃2 (𝑠)/𝑇(𝑠), for the rotational system shown in figure The rod is supported by bearings at either end and is undergoing torsion A torque is applied at the left, and the displacement is measured at the right System Dynamics and Control 2.104 Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions Next, draw a free-body diagram of 𝐽1 𝐽2, using superposition Torques on 𝐽1 due only to the motion of 𝐽1 Torques on 𝐽1 due only to the motion of 𝐽2 Final free-body diagram for 𝐽1 𝐽1 𝑠 + 𝐷1 𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) (2.127.a) Solution First, obtain the schematic from the physical system Torques on 𝐽2 due only Final free-body diagram for 𝐽2 to the motion of 𝐽1 + (𝐽2 𝑠 + 𝐷2 𝑠 + 𝐾)𝜃2 𝑠 = (2.127.b) Torques on 𝐽2 due only to the motion of 𝐽2 −𝐾1 𝜃1 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.105 Nguyen Tan Tien Modeling in Frequency Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.106 Nguyen Tan Tien Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions 𝐽1 𝑠 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠)(2.127.a) −𝐾1 𝜃1 𝑠 + (𝐽2 𝑠 + 𝐷2 𝑠 + 𝐾)𝜃2 𝑠 = (2.127.b) The solution for 𝜃2 𝐽1 𝑠 + 𝐷1 𝑠 + 𝐾 𝑇 −𝐾 = 𝐾𝑇 𝜃2 = ∆ ∆ where 𝐽 𝑠 + 𝐷1 𝑠 + 𝐾 −𝐾 ∆= −𝐾 𝐽2 𝑠 + 𝐷2𝑠 + 𝐾 𝜃2 (𝑠) 𝐾 ⟹𝐺 𝑠 = = 𝑇(𝑠) ∆ §6.Rotational Mechanical System Transfer Functions Note HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.107 Nguyen Tan Tien Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions - Ex.2.20 Equations of Motion by Inspection Write the Laplace transform of the equations of motion for the system shown in the figure Solution The Laplace transform of the equations of motion of 𝐽1 + 𝐽1𝑠 + 𝐷1𝑠 + 𝐾 𝜃1 − 𝐾𝜃2 − 0𝜃3 = 𝑇(𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering 𝐽1 𝑠 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) (2.127.a) −𝐾1 𝜃1 𝑠 + (𝐽2𝑠 + 𝐷2 𝑠 + 𝐾)𝜃2 𝑠 = (2.127.b) System Dynamics and Control 2.108 Nguyen Tan Tien Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions The Laplace transform of the equations of motion of 𝐽2 (2.131.a) Nguyen Tan Tien −𝐾𝜃1 + 𝐽2 𝑠 + 𝐷2 𝑠 + 𝐾 𝜃2 − 𝐷2 𝑠𝜃3 = HCM City Univ of Technology, Faculty of Mechanical Engineering (2.131.b) Nguyen Tan Tien 18 1/11/2016 System Dynamics and Control 2.109 Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions System Dynamics and Control 2.110 Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions Skill-Assessment Ex.2.9 Problem Find the transfer function 𝜃2 (𝑠) 𝐺 𝑠 = 𝑇(𝑠) Solution The equations of motion The Laplace transform of the equations of motion of 𝐽3 + 𝑠 + 𝑠 + 𝜃1 𝑠 −0𝜃1 − 𝐷2𝑠𝜃2 + 𝐽3𝑠 + 𝐷3 𝑠 + 𝐷2 𝑠 𝜃3 = (2.131.c) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.111 Nguyen Tan Tien Modeling in Frequency Domain §6.Rotational Mechanical System Transfer Functions The equations of motion 𝑠 + 𝑠 + 𝜃1 𝑠 − (𝑠 + 1)𝜃2 𝑠 = 𝑇(𝑠) − 𝑠 + 𝜃1 𝑠 + (2𝑠 + 2)𝜃2 𝑠 = Solving for 𝜃2 (𝑠) 𝑠2 + 𝑠 + 𝑇 (𝑠 + 1)𝑇 − 𝑠+1 𝜃2 = = 2𝑠 + 3𝑠 + 2𝑠 + 𝑠 + 𝑠 + −(𝑠 + 1) − 𝑠+1 2𝑠 + 𝜃2 (𝑠) 𝑠+1 ⟹𝐺 𝑠 = = 𝑇(𝑠) 2𝑠 + 3𝑠 + 2𝑠 + HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.113 Nguyen Tan Tien Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears − 𝑠 + 𝜃1 𝑠 + (2𝑠 + 2)𝜃2 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.112 Nguyen Tan Tien Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears Kinematic relationship 𝜃2 𝑟1 𝑁1 = = 𝜃1 𝑟2 𝑁2 Power on gears 𝑇1 𝜃1 = 𝑇2 𝜃2 The ratio of torques on two gears 𝑇2 𝜃1 𝑁2 = = 𝑇1 𝜃2 𝑁1 𝜃1 , 𝜃2 : rotation angles of gear and 2, 𝑟𝑎𝑑 𝑟1, 𝑟2 : radius of gear and 2, 𝑚 𝑁1 , 𝑁2 : number of teeth of gear and 𝑇1, 𝑇2 : torques on gear and 2, 𝑁𝑚 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.114 Nguyen Tan Tien Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears a.rotational system driven by b.equivalent system at the output after reflection of input torque gears a.rotational system driven by b.equivalent system at the output after reflection of input torque gears What happens to mechanical impedances that are driven by gears? (a) : gears driving a rotational inertia, spring, and viscous damper (b) : an equivalent system at 𝜃1 without the gears Can the mechanical impedances be reflected from the output to the input, thereby eliminating the gears? HCM City Univ of Technology, Faculty of Mechanical Engineering − (𝑠 + 1)𝜃2 𝑠 = 𝑇(𝑠) Nguyen Tan Tien 𝑇1 can be reflected to the output by multiplying by 𝑁2 /𝑁1 𝑁2 𝐽𝑠 + 𝐷𝑠 + 𝐾 𝜃2 𝑠 = 𝑇1 (𝑠) (2.131) 𝑁1 Convert 𝜃2 (𝑠) into an equivalent 𝜃1 (𝑠), so that 𝑁1 𝑁2 𝐽𝑠 + 𝐷𝑠 + 𝐾 𝜃 𝑠 = 𝑇1 (𝑠) (2.132) 𝑁2 𝑁1 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 19 1/11/2016 System Dynamics and Control 2.115 Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears 𝐽𝑠 + 𝐷𝑠 + 𝐾 𝜃2 𝑠 = 𝑇1 (𝑠)(𝑁2 /𝑁1 ) 𝐽𝑠 + 𝐷𝑠 + 𝐾 (𝑁2 /𝑁1 )𝜃1 𝑠 = 𝑇1 (𝑠)(𝑁2 /𝑁1 ) 𝑁1 𝑁2 𝑠2 + 𝐷 𝑁1 𝑁2 𝑠+𝐾 𝑁1 𝑁2 2.116 Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears Generalizing the results Rotational mechanical impedances can be reflected through gear trains by multiplying the mechanical impedance by the ratio a.rotational system driven by b.equivalent system at the output c.equivalent system at the input after reflection of input torque after reflection of impedances gears ⟹ 𝐽 System Dynamics and Control (2.131) (2.132) where the impedance to be reflected is attached to the source shaft and is being reflected to the destination shaft 𝜃1 𝑠 = 𝑇1 (𝑠) (2.133) Thus, the load can be thought of as having been reflected from the output to the input HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.117 Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears - Ex.2.21 Transfer Function - System with Lossless Gears Find the transfer function, 𝜃2 (𝑠)/𝑇1 (𝑠), for the system a.rotational mechanicalsystem with gears b.system after reflection of torques and impedances to the output shaft Solution Reflect the impedances (𝐽1 and 𝐷1) and torque (𝑇1) on the input shaft to the output, where the impedances are reflected by (𝑁1 /𝑁2 )2 and the torque is reflected by (𝑁1 /𝑁2 ) The equation of motion can now be written as 𝐽𝑒 𝑠 + 𝐷𝑒 𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 𝑠 (𝑁2 /𝑁1 ) (2.139) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.119 Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears - In order to eliminate gears with large radii, a gear train is used to implement large gear ratios by cascading smaller gear ratios 𝜃4 = 𝑁1 𝑁3 𝑁5 𝜃 𝑁2 𝑁4 𝑁6 System Dynamics and Control Nguyen Tan Tien 2.118 Nguyen Tan Tien Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears a.rotational mechanicalsystem with gears b.system after reflection of torques and impedances to the output shaft 𝐽𝑒 𝑠 + 𝐷𝑒 𝑠 + 𝐾𝑒 𝜃2 𝑠 = 𝑇1 𝑠 (𝑁2 /𝑁1 ) (2.139) where, 𝐽𝑒 = 𝐽1 (𝑁2 /𝑁1 )2 +𝐽2, 𝐷𝑒 = 𝐷1(𝑁2 /𝑁1 )2 +𝐷2, 𝐾𝑒 = 𝐾2 Solving for 𝐺(𝑠) 𝜃2 (𝑠) 𝑁2 /𝑁1 𝐺 𝑠 = = 𝑇1 (𝑠) 𝐽𝑒 𝑠 + 𝐷𝑒 𝑠 + 𝐾𝑒 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.120 Nguyen Tan Tien Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears - Ex.2.22 Transfer Function – Gears with Loss Find the transfer function, 𝜃1 (𝑠)/𝑇1(𝑠), for the system a.system using a gear train - For gear trains, the equivalent gear ratio is the product of the individual gear ratios HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering b.equivalent system at the input Solution Reflect all of the impedances to the input shaft, 𝜃1 The equation of motion can now be written as 𝐽𝑒 𝑠 + 𝐷𝑒 𝑠 𝜃1 𝑠 = 𝑇1 𝑠 The transfer function 𝐺 𝑠 = 𝜃1 𝑠 /𝑇1 𝑠 = 1/(𝐽𝑒 𝑠 + 𝐷𝑒 𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 20 1/11/2016 System Dynamics and Control 2.121 Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears Skill-Assessment Ex.2.10 Problem Find the TF 𝜃2 (𝑠) 𝐺 𝑠 = 𝑇(𝑠) System Dynamics and Control 2.122 Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears The equation of motion Solution Transforming the network to one without gears by reflecting the 4𝑁𝑚/𝑟𝑎𝑑 spring to the left and multiplying by (25/50)2 25 4[𝑁𝑚/𝑟𝑎𝑑] × 50 + 𝑠 + 𝑠 𝜃1 𝑠 − 𝑠𝜃𝑎 𝑠 = 𝑇(𝑠) = 1[𝑁𝑚/𝑟𝑎𝑑] −𝑠𝜃1 𝑠 + (𝑠 + 1)𝜃𝑎 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.123 Nguyen Tan Tien Modeling in Frequency Domain §7.Transfer Functions for Systems with Gears The equation of motion 𝑠 + 𝑠 𝜃1 𝑠 − 𝑠𝜃𝑎 𝑠 = 𝑇(𝑠) −𝑠𝜃1 𝑠 + (𝑠 + 1)𝜃𝑎 𝑠 = Solving for 𝜃𝑎 (𝑠) 𝑠2 + 𝑠 𝑇 = 𝑠𝑇(𝑠) 𝜃𝑎 𝑠 = −𝑠 𝑠 +𝑠 −𝑠 𝑠3 + 𝑠2 + 𝑠 −𝑠 𝑠+1 𝜃𝑎 𝑠 ⟹ = 𝑇(𝑠) 𝑠 + 𝑠 + The transfer function 𝜃𝑎 𝑠 𝜃2 𝑠 1/2 =2 = 𝑇(𝑠) 𝑇(𝑠) 𝑠 +𝑠+1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.125 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.124 Nguyen Tan Tien Modeling in Frequency Domain §8.Electromechanical System Transfer Functions NASA flight simulator robot arm with electromechanical control system components Nguyen Tan Tien Modeling in Frequency Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.126 Nguyen Tan Tien Modeling in Frequency Domain §8.Electromechanical System Transfer Functions - A motor is an electromechanical component that yields a displacement output for a voltage input, that is, a mechanical output generated by an electrical input - Derive the transfer function for the armature-controlled dc servomotor (Mablekos, 1980) • Fixed field: a magnetic field is developed by stationary permanent magnets or a stationary electromagnet • Armature: a rotating circuit, through which current 𝑖𝑎 (𝑡) flows, passes through this magnetic field at right angles and feels a force 𝐹 = 𝐵𝑙𝑖𝑎 (𝑡) 𝐵 : the magnetic field strength 𝑙 : the length of the conductor §8.Electromechanical System Transfer Functions • A conductor moving at right angles to a magnetic field generates a voltage at the terminals of the conductor equal to 𝑒 = 𝐵𝑙𝑣 𝑒 : the voltage 𝑣 : the velocity of the conductor normal to the magnetic field • The current-carrying armature is rotating in a magnetic field, its voltage is proportional to speed HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 𝑣𝑏 𝑡 = 𝐾𝑏 𝜃𝑚 (𝑡) (2.144) 𝑣𝑏 𝑡 : the back electromotive force (back emf) 𝐾𝑏 : a constant of proportionality called the back emf constant 𝜃𝑚(𝑡): the angular velocity of the motor Nguyen Tan Tien 21 1/11/2016 System Dynamics and Control 2.127 Modeling in Frequency Domain System Dynamics and Control 2.128 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions • Taking the Laplace transform 𝑉𝑏 (𝑠) = 𝐾𝑏 𝑠𝜃𝑚 (𝑠) (2.145) • The relationship between the armature current, 𝑖𝑎 (𝑡) , the applied armature voltage, 𝑒𝑎 (𝑡), and the back emf, 𝑣𝑏 (𝑡) 𝑅𝑎 𝐼𝑎 𝑠 + 𝐿𝑎 𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146) • The torque developed by the motor is proportional to the armature current 𝑇𝑚 𝑠 = 𝐾𝑡 𝐼𝑎 𝑠 (2.147) 𝑇𝑚 : the torque developed by the motor 𝐾𝑡 : the motor torque constant, which depends on the motor and magnetic field characteristics Đ8.Electromechanical System Transfer Functions ã Rearranging Eq.(2.147) 𝐼𝑎 (𝑠) = 𝑇𝑚 (𝑠) (2.148) 𝐾𝑡 • To find the TF of the motor, first substitute Eqs (2.145) and (2.148) into (2.146), yielding (𝑅𝑎 + 𝐿𝑎 𝑠)𝑇𝑚 (𝑠) (2.149) + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 𝑠 𝐾𝑡 • Then, find 𝑇𝑚 (𝑠) in terms of 𝜃𝑚 (𝑠), separate the input and output variables and obtain the transfer function, 𝜃𝑚 (𝑠)/𝐸𝑎 (𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑉𝑏 (𝑠) = 𝐾𝑏 𝑠𝜃𝑚 (𝑠) (2.145), System Dynamics and Control 2.129 Nguyen Tan Tien Modeling in Frequency Domain §8.Electromechanical System Transfer Functions • A typical equivalent mechanical loading on a motor 𝐽𝑚 : the equivalent inertia at the armature and includes both the armature inertia and, the load inertia reflected to the armature 𝐷𝑚 : the equivalent viscous damping at the armature and includes both the armature viscous damping and, the load viscous damping reflected to the armature 𝑇𝑚 𝑠 = (𝐽𝑚 𝑠 + 𝐷𝑚 𝑠)𝜃𝑚 (𝑠) (2.150) • Substituting Eq.(2.150) into Eq.(2.149) (𝑅𝑎 + 𝐿𝑎𝑠)(𝐽𝑚𝑠2 + 𝐷𝑚𝑠)𝜃𝑚(𝑠) + 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎(𝑠) (2.151) 𝐾𝑡 (𝑅𝑎 +𝐿𝑎 𝑠)𝑇𝑚 (𝑠) 𝐾𝑡 + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 𝑠 (2.149) 2.131 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions • Let us first discuss the mechanical constants, 𝐽𝑚 and 𝐷𝑚 Consider the figure: a motor with inertia 𝐽𝑎 and damping 𝐷𝑎 at the armature driving a load consisting of inertia 𝐽𝐿 and damping 𝐷𝐿 Assuming that all inertia and damping values shown are known, 𝐽𝐿 and 𝐷𝐿 can be reflected back to the armature as some equivalent inertia and damping to be added to 𝐽𝑎 and 𝐷𝑎 , respectively Thus, the equivalent inertia, 𝐽𝑚 , and equivalent damping, 𝐷𝑚 , at the armature are 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 𝑁1 𝑁2 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 𝑁1 𝑁2 (2.155.a) HCM City Univ of Technology, Faculty of Mechanical Engineering Modeling in Frequency Domain • After simplification 𝐾𝑡 𝜃𝑚 (𝑠) 𝑅𝑎 𝐽𝑚 = 𝐸𝑎 (𝑠) 𝑠 𝑠 + 𝐷 + 𝐾𝑡 𝐾 𝐽𝑚 𝑚 𝑅𝑎 𝑏 • The form of Eq.(2.153) 𝜃𝑚 (𝑠) 𝐾 = 𝐸𝑎 (𝑠) 𝑠(𝑠 + 𝛼) 𝐾𝑡 Nguyen Tan Tien 2.130 Nguyen Tan Tien Đ8.Electromechanical System Transfer Functions ã Assume that the armature inductance, 𝐿𝑎 , is small compared to the armature resistance, 𝑅𝑎 , which is usual for a dc motor, Eq (2.151) becomes 𝑅𝑎 𝐽 𝑠 + 𝐷𝑚 + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 (𝑠) (2.152) 𝐾𝑡 𝑚 (𝑅𝑎 +𝐿𝑎𝑠)(𝐽𝑚 𝑠2+𝐷𝑚𝑠)𝜃𝑚 (𝑠) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control System Dynamics and Control 𝑅𝑎 𝐼𝑎 𝑠 + 𝐿𝑎 𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146) (2.153) (2.154) + 𝐾𝑏𝑠𝜃𝑚 𝑠 = 𝐸𝑎(𝑠) (2.151) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.132 Nguyen Tan Tien Modeling in Frequency Domain Đ8.Electromechanical System Transfer Functions ã Substituting Eqs.(2.145), (2.148) into Eq (2.146), with 𝐿𝑎 = 𝑅𝑎 𝑇 𝑠 + 𝐾𝑏 𝑠𝜃𝑚 𝑠 = 𝐸𝑎 𝑠 (2.156) 𝐾𝑡 𝑚 Taking the inverse Laplace transform 𝑅𝑎 𝑇 𝑡 + 𝐾𝑏 𝜔𝑚 𝑡 = 𝑒𝑎 𝑡 (2.157) 𝐾𝑡 𝑚 • When the motor is operating at steady state with a dc voltage input 𝑅𝑎 𝑇 + 𝐾𝑏 𝜔𝑚 = 𝑒𝑎 (2.158) 𝐾𝑡 𝑚 𝐾𝑏 𝐾𝑡 𝐾𝑡 ⟹ 𝑇𝑚 = − 𝜔 + 𝑒 (2.159) 𝑅𝑎 𝑚 𝑅𝑎 𝑎 (2.155.b) 𝑉𝑏 (𝑠) = 𝐾𝑏 𝑠𝜃𝑚 (𝑠) (2.145), 𝑅𝑎 𝐼𝑎 𝑠 + 𝐿𝑎 𝑠𝐼𝑎 𝑠 + 𝑉𝑏 𝑠 = 𝐸𝑎 𝑠 (2.146), 𝐼𝑎 (𝑠) = 𝐾 𝑇𝑚 (𝑠) (2.148) 𝑡 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 22 1/11/2016 System Dynamics and Control 2.133 Modeling in Frequency Domain System Dynamics and Control 2.134 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions The stall torque 𝐾𝑡 𝑇𝑠𝑡𝑎𝑙𝑙 = 𝑒 (2.160) 𝑅𝑎 𝑎 The no-load speed 𝑒𝑎 (2.161) 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 = 𝐾𝑏 The electrical constants of the motor 𝐾𝑡 𝑇𝑠𝑡𝑎𝑙𝑙 (2.162) = 𝑅𝑎 𝑒𝑎 𝑒𝑎 (2.163) 𝐾𝑏 = 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 The electrical constants, 𝐾𝑡 /𝑅𝑎 and 𝐾𝑏 , can be found from a dynamometer test of the motor, which would yield 𝑇𝑠𝑡𝑎𝑙𝑙 and 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 for a given 𝑒𝑎 §8.Electromechanical System Transfer Functions - Ex.2.23 Transfer Function-DC Motor and Load 𝜃𝐿 𝑠 Given the system and torque-speed curve, find the TF, 𝐸𝑎 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.135 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions Solution Find the mechanical constants 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 𝑁1 𝑁2 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 𝑁1 𝑁2 = + 700 × = + 800 × HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 100 1000 2.137 System Dynamics and Control Nguyen Tan Tien 2.136 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions Find the electrical constants from the torque-speed curve 𝑇𝑠𝑡𝑎𝑙𝑙 = 500, 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 = 50, 𝑒𝑎 = 100 100 1000 𝐾𝑡 𝑇𝑠𝑡𝑎𝑙𝑙 500 = = =5 𝑅𝑎 𝑒𝑎 100 𝑒𝑎 100 𝐾𝑏 = = =2 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 50 = 12 = 10 Nguyen Tan Tien Modeling in Frequency Domain HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control Nguyen Tan Tien 2.138 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions The transfer function 𝜃𝑚 (𝑠)/𝐸𝑎 (𝑠) 𝐾𝑡 𝜃𝑚 (𝑠) 𝑅𝑎 𝐽𝑚 = 𝐸𝑎 (𝑠) 𝑠 𝑠 + 𝐷 + 𝐾𝑡 𝐾 𝐽𝑚 𝑚 𝑅𝑎 𝑏 5× 12 = 𝑠 𝑠+ × 10 + × 12 0.417 = 𝑠(𝑠 + 1.667) The transfer function 𝜃𝐿 (𝑠)/𝐸𝑎 (𝑠) 𝑁1 100 𝜃𝐿 (𝑠) 𝜃𝑚 (𝑠) 𝑁2 0.417 × 1000 0.0417 = = = 𝐸𝑎 (𝑠) 𝐸𝑎 (𝑠) 𝑠(𝑠 + 1.667) 𝑠(𝑠 + 1.667) §8.Electromechanical System Transfer Functions Skill-Assessment Ex.2.11 Problem Find the TF, 𝐺 𝑠 = 𝜃𝐿 (𝑠)/𝐸𝑠 (𝑠), for the motor and load system The torque-speed curve is given by 𝑇𝑚 = − 8𝜔𝑚 + 200 when the input voltage is 100𝑣𝑜𝑙𝑡𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Solution Find the mechanical constants 𝐽𝑚 = 𝐽𝑎 + 𝐽𝐿 𝑁1 𝑁3 𝑁2 𝑁4 𝐷𝑚 = 𝐷𝑎 + 𝐷𝐿 𝑁1 𝑁3 𝑁2 𝑁4 = + 400 × 20 25 × 100 100 = + 800 × =2 20 25 × 100 100 =7 Nguyen Tan Tien 23 1/11/2016 System Dynamics and Control 2.139 Modeling in Frequency Domain §8.Electromechanical System Transfer Functions Find the electrical constants from the torque-speed eq 𝜔𝑚 = ⟹ 𝑇𝑚 = 200 𝑇𝑚 = ⟹ 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 = 200/8 = 25 𝐾𝑡 𝑇𝑠𝑡𝑎𝑙𝑙 200 = = =2 𝑅𝑎 𝐸𝑎 100 𝐸𝑎 100 𝐾𝑏 = = =4 𝜔𝑛𝑜−𝑙𝑜𝑎𝑑 25 𝑇𝑚 = −8𝜔𝑚 + 200 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.141 Nguyen Tan Tien Modeling in Frequency Domain System Dynamics and Control 2.140 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.142 §9.Electric Circuit Analogs - Electric circuit analog: an electric circuit that is analogous to a system from another discipline - In the commonality of systems from the various disciplines, the mechanical systems can be represented by equivalent electric circuits - Analogs can be obtained by comparing the describing equations, such as the equations of motion of a mechanical system, with either electrical mesh or nodal equations • When compared with mesh equations, the resulting electrical circuit is called a series analog • When compared with nodal equations, the resulting electrical circuit is called a parallel analog §9.Electric Circuit Analogs Series Analog HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.143 Nguyen Tan Tien Modeling in Frequency Domain §9.Electric Circuit Analogs Parameters for series analog mass =𝑀 ⟹ viscous damper = 𝑓𝑣 ⟹ spring =𝐾 ⟹ applied force = 𝑓(𝑡) ⟹ velocity = 𝑣(𝑡) ⟹ Modeling in Frequency Domain §8.Electromechanical System Transfer Functions Substituting all values into the motor transfer function 𝐾𝑡 𝜃𝑚 (𝑠) 𝑅𝑎 𝐽𝑚 = 𝐸𝑎 (𝑠) 𝑠 𝑠 + 𝐷 + 𝐾𝑡 𝐾 𝐽𝑚 𝑚 𝑅𝑎 𝑏 2× = 𝑠 𝑠+ 7+2×4 = 𝑠(𝑠 + 7.5) The transfer function 𝜃𝐿 (𝑠)/𝐸𝑎 (𝑠) 𝑁1 𝑁3 20 25 𝜃𝐿 (𝑠) 𝜃𝑚 (𝑠) 𝑁2 𝑁4 100 × 100 0.05 = = = 𝐸𝑎 (𝑠) 𝐸𝑎 (𝑠) 𝑠(𝑠 + 7.5) 𝑠(𝑠 + 7.5) Nguyen Tan Tien Modeling in Frequency Domain Consider the translational mechanical system, the equation of motion 𝑀𝑠 + 𝑓𝑣 𝑠 + 𝐾 𝑀𝑠 + 𝑓𝑣 𝑠 + 𝐾 𝑋 𝑠 = 𝐹 𝑠 = 𝑠𝑋 𝑠 𝑠 𝐾 ⟹ 𝑀𝑠 + 𝑓𝑣 + 𝑉 𝑠 = 𝐹(𝑠) 𝑠 Kirchhoff’s mesh equation for the simple series RLC network 𝐿𝑠 + 𝑅 + 𝐼 𝑠 = 𝐸(𝑠) 𝐶𝑠 System Dynamics and Control 2.144 Nguyen Tan Tien Modeling in Frequency Domain §9.Electric Circuit Analogs - Ex.2.24 Converting a Mechanical System to a Series Analog Draw a series analog for the mechanical system inductor resistor capacitor voltage source mesh current HCM City Univ of Technology, Faculty of Mechanical Engineering 𝐿 = 𝑀 henries 𝑅 = 𝑓𝑣 ohms 𝐶 = 1/𝐾 farads 𝑒 𝑡 = 𝑓(𝑡) 𝑖 𝑡 = 𝑣(𝑡) Nguyen Tan Tien Solution The equations of motion with 𝑋(𝑠) → 𝑉(𝑠) 𝐾1 + 𝐾2 𝐾2 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 + 𝑉1 𝑠 − 𝑓𝑣3 + 𝑉2 𝑠 = 𝐹(𝑠) 𝑠 𝑠 𝐾2 𝐾2 + 𝐾3 − 𝑓𝑣3 + 𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 + 𝑉2 𝑠 = 𝑠 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 24 1/11/2016 System Dynamics and Control 2.145 Modeling in Frequency Domain §9.Electric Circuit Analogs Coefficients represent sums of electrical impedance Mechanical impedances associated with 𝑀1 form the first mesh, where impedances between the two masses are common to the two loops Impedances associated with 𝑀2 form the second mesh 𝑣1 (𝑡) and 𝑣2 (𝑡)are the velocities of 𝑀1 and 𝑀2 , respectively System Dynamics and Control 2.146 Modeling in Frequency Domain §9.Electric Circuit Analogs Parallel Analog Consider the translational mechanical system, the equation of motion 𝐾 𝑀𝑠 + 𝑓𝑣 + 𝑉 𝑠 = 𝐹(𝑠) 𝑠 + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 + − 𝑓𝑣3 + 𝐾2 𝑠 𝐾1 +𝐾2 𝑠 𝑉1 𝑠 − 𝑓𝑣3 + 𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 + 𝐾2 𝑠 𝐾2 +𝐾3 𝑠 Kirchhoff’s nodal equation for the simple parallel RLC network 1 𝐶𝑠 + + 𝐸 𝑠 = 𝐼(𝑠) 𝑅𝑠 𝐿𝑠 𝑉2 𝑠 = 𝐹(𝑠) 𝑉2 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.147 Nguyen Tan Tien Modeling in Frequency Domain §9.Electric Circuit Analogs System Dynamics and Control 2.148 Nguyen Tan Tien Modeling in Frequency Domain §9.Electric Circuit Analogs - Ex.2.25 Converting a Mechanical System to a Parallel Analog Draw a parallel analog for the mechanical system Parameters for parallel analog mass =𝑀 ⟹ viscous damper = 𝑓𝑣 ⟹ spring =𝐾 ⟹ applied force = 𝑓(𝑡) ⟹ velocity = 𝑣(𝑡) ⟹ capacitor resistor inductor current source node voltage HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control HCM City Univ of Technology, Faculty of Mechanical Engineering 2.149 𝐶 = 𝑀 farads 𝑅 = 1/𝑓𝑣 ohms 𝐿 = 1/𝐾 henries 𝑖(𝑡) = 𝑓(𝑡) 𝑒(𝑡) = 𝑣(𝑡) Nguyen Tan Tien Modeling in Frequency Domain §9.Electric Circuit Analogs Coefficients represent sums of electrical admittances Admittances associated with 𝑀1 form the elements connected to the first node, where mechanical admittances between the two masses are common to the two nodes Mechanical admittances associated with 𝑀2 form the elements connected to the second node 𝑣1 (𝑡) and 𝑣2 (𝑡) are the velocities of 𝑀1 and 𝑀2 , respectively Solution The equations of motion with 𝑋(𝑠) → 𝑉(𝑠) 𝐾1 + 𝐾2 𝐾2 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 + 𝑉1 𝑠 − 𝑓𝑣3 + 𝑉2 𝑠 = 𝐹(𝑠) 𝑠 𝑠 𝐾2 𝐾2 + 𝐾3 − 𝑓𝑣3 + 𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 + 𝑉2 𝑠 = 𝑠 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics and Control 2.150 Nguyen Tan Tien Modeling in Frequency Domain §9.Electric Circuit Analogs Skill-Assessment Ex.2.12 Problem Draw a series and parallel analog for the rotational mechanical system Solution The equations of motion + 𝐽1 𝑠 + 𝐷1𝑠 + 𝐾 𝜃1 𝑠 − 𝐾𝜃2 𝑠 = 𝑇(𝑠) −𝐾𝜃1 𝑠 + 𝐽2𝑠 + 𝐷2 𝑠 + 𝐾 𝜃2 𝑠 = + 𝑀1 𝑠 + 𝑓𝑣1 + 𝑓𝑣3 + − 𝐾 𝑓𝑣3 + 𝑠2 𝐾1 +𝐾2 𝑠 𝐾2 𝑠 𝐾2 +𝐾3 + 𝑠 𝑉1 𝑠 − 𝑓𝑣3 + 𝑉1 𝑠 + 𝑀2 𝑠 + 𝑓𝑣2 + 𝑓𝑣3 𝑉2 𝑠 = 𝐹(𝑠) 𝑉2 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 25 ... 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