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# Ch.02 modeling in frequency domain

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Ch.02 modeling in frequency domain

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### Ch.02 modeling in frequency domain

1. 1. 1/11/2016 1 02. Modeling in Frequency Domain System Dynamics and Control 2.01 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Chapter Objectives After completing this chapter, the student will be able to β’ Find the Laplace transform of time functions and the inverse Laplace transform β’ Find the transfer function (TF) from a differential equation and solve the differential equation using the transfer function β’ Find the transfer function for linear, time-invariant electrical networks β’ Find the TF for linear, time-invarianttranslational mechanical systems β’ Find the TF for linear, time-invariant rotational mechanical systems β’ Find the TF for gear systems with no loss and for gear systems with loss β’ Find the TF for linear, time-invariant electromechanical systems β’ Produce analogous electrical and mechanical circuits β’ Linearize a nonlinear system in order to find the TF System Dynamics and Control 2.02 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§1.Introduction - Mathematical models from schematics of physical systems β’ transfer functions in the frequency domain β’ state equations in the time domain System Dynamics and Control 2.03 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review - Transforms: a mathematical conversion from one way of thinking to another to make a problem easier to solve - The Laplace transform the problem in time-domain to problem in π -domain, then applying the solution in π -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, β¦ System Dynamics and Control 2.04 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review - The Laplace transform of the function π(π‘) for π‘ > 0 is defined by the following relationship π  : complex frequency variable, π  = π + ππ with π , π are real numbers, π  β πΆ for which makes πΉ π  convergent β : Laplace transform πΉ(π ): a complex-valued function of complex numbers - The inverse Laplace transform of the function πΉ(π ) for π‘ > 0 is defined by the following relationship π’(π‘) : the unit step function, π’ π‘ = 1 ππ π‘ > 0 0 ππ π‘ < 0 System Dynamics and Control 2.05 Modeling in Frequency Domain πΉ π  = β π(π‘) = 0β +β π(π‘)πβπ π‘ ππ‘ π π‘ = ββ1 πΉ π  = 1 2ππ πβπβ π+πβ πΉ π  π π π‘ ππ  = π π‘ π’(π‘) (2.1) (2.2) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review - The Laplace transform table System Dynamics and Control 2.06 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2. 2. 1/11/2016 2 Β§2.Laplace Transform Review - Ex.2.1 Laplace Transform of a Time Function Find the Laplace transform of π π‘ = π΄πβππ‘ π’(π‘) Solution πΉ π  = 0 β π(π‘)πβπ π‘ ππ‘ = 0 β π΄πβππ‘ πβπ π‘ ππ‘ = π΄ 0 β πβ(π+π )π‘ ππ‘ = β π΄ π  + π πβ(π+π )π‘ 0 β βΉ πΉ π  = π΄ π  + π System Dynamics and Control 2.07 Modeling in Frequency Domain (2.3) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review System Dynamics and Control 2.08 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien β π‘π’ π‘ = 1 π 2 (Table 2.1 β 3) β πβππ‘ π(π‘) = πΉ(π  + π) (Table 2.2 β 4) Β§2.Laplace Transform Review - Ex.2.2 Inverse Laplace Transform Find the inverse Laplace transform of πΉ1 π  = 1/(π  + 3)2 Solution π π‘ = ββ1 1 π 2 = π‘π’(π‘) π1 π‘ = ββ1 1 (π  + 3)2 = πβ3π‘ π π‘ βΉ π1 π‘ = πβ3π‘ π‘π’(π‘) System Dynamics and Control 2.09 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien β πΏ π‘ = 1 (Table 2.1 β 1) β ππ(π‘) ππ‘ = π πΉ(π ) β β ππΏ(π‘) ππ‘ = π  (Table 2.2 β 7) Β§2.Laplace Transform Review Partial-Fraction Expansion πΉ1 π  = π 3 + 2π 2 + 6π  + 7 π 2 + π  + 5 = (π  + 1) + 2 π 2 + π  + 5 βΉ π1 π‘ = ππΏ(π‘) ππ‘ + πΏ π‘ + ββ1 2 π 2 + π  + 5 Using partial-fraction expansion to expand function like πΉ(π ) into a sum of terms and then find the inverse Laplace transform for each term System Dynamics and Control 2.10 Modeling in Frequency Domain πΉ(π ) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review Case 1. Roots of the Denominator of πΉ(π ) are Real and Distinct πΉ π  = 2 (π  + 1)(π  + 2) = πΎ1 π  + 1 + πΎ2 π  + 2 lim π ββ1 [(2.8) Γ (π  + 1)] βΉ lim π ββ1 2 π  + 2 = lim π ββ1 πΎ1 + (π  + 1)πΎ2 π  + 2 βΉ πΎ1 = 2 lim π ββ2 [(2.8) Γ (π  + 2)] βΉ lim π ββ2 2 π  + 1 = lim π ββ2 (π  + 2)πΎ1 π  + 1 + πΎ2 βΉ πΎ2 = β2 βΉ πΉ π  = 2 π  + 1 β 2 π  + 2 βΉ π π‘ = 2πβπ‘ β 2πβ2π‘ π’(π‘) System Dynamics and Control 2.11 Modeling in Frequency Domain (2.8) (2.10) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review In general, given an πΉ(π ) whose denominator has real and distinct roots, a partial-fraction expansion πΉ π  = π(π ) π·(π ) = π(π ) π  + π1 π  + π2 β¦ π  + ππ β¦ (π  + π π) = πΎ1 π  + π1 + πΎ2 π  + π2 + β― + πΎπ π  + ππ + β― + πΎπ π  + π π To find πΎπ β’ multiply (2.11) by π  + ππ β’ let π  approach βππ System Dynamics and Control 2.12 Modeling in Frequency Domain (2.11) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
3. 3. 1/11/2016 3 β ππ ππ‘ = π πΉ π  β π(0β) (Table 2.2 β 7) β π2 π ππ‘2 = π 2 πΉ π  β π π(0β) β πβ²(0β) (Table 2.2 β 8) Β§2.Laplace Transform Review - Ex.2.3 Laplace Transform Solution of a Differential Equation Given the following differential equation, solve for π¦(π‘) if all initial conditions are zero. Use the Laplace transform π2 π¦ ππ‘2 + 12 ππ¦ ππ‘ + 32π¦ = 32π’(π‘) Solution π 2 π π  + 12π π π  + 32π π  = 32 π  βΉ π π  = 32 π  π 2 + 12π  + 32 = 32 π (π  + 4)(π  + 8) = πΎ1 π  + πΎ2 π  + 4 + πΎ3 π  + 8 System Dynamics and Control 2.13 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review π π  = 32 π (π  + 4)(π  + 8) = πΎ1 π  + πΎ2 π  + 4 + πΎ3 π  + 8 Evaluate the residue πΎπ πΎ1 = 32 (π  + 4)(π  + 8) π β0 = 1 πΎ2 = 32 π (π  + 8) π ββ4 = β2 πΎ2 = 32 π (π  + 4) π ββ8 = 1 βΉ π π  = 1 π  β 2 π  + 4 + 1 π  + 8 Hence π¦ π‘ = 1 β 2πβ4π‘ + πβ8π‘ π’(π‘) System Dynamics and Control 2.14 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review π π  = 32 π  π 2 + 12π  + 32 = 1 π  β 2 π  + 4 + 1 π  + 8 π¦ π‘ = 1 β 2πβ4π‘ + πβ8π‘ π’(π‘) (2.20) The π’(π‘) in (2.20) shows that the response is zero until π‘ = 0 Unless otherwise specified, all inputs to systems in the text will not start until π‘ = 0. Thus, output responses will also be zero until π‘ = 0 For convenience, the π’(π‘) notation will be eliminated from now on. Accordingly, the output response π¦ π‘ = 1 β 2πβ4π‘ + πβ8π‘ (2.21) System Dynamics and Control 2.15 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review Run ch2p1 through ch2p8 in Appendix B Learn how to use MATLAB to β’ represent polynomials β’ find roots of polynomials β’ multiply polynomials, and β’ find partial-fraction expansions System Dynamics and Control 2.16 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 2.17 Modeling in Frequency Domain Β§2.Laplace Transform Review π π  = 32 π 3 + 12π 2 + 32π  = 1 π  β 2 π  + 4 + 1 π  + 8 Matlab [r,p,k] = residue([32],[1,12,32,0]) Result r = [1, β2, 1], p = [β8, β4, 0], k = [ ] π π  = 0 π + 1 π1 1 π  β (β8 π1 ) + (β2 π2 ) 1 π  β (β4 π2 ) + 1 π3 1 π  β (0 π3 ) = 1 π  + 8 β 2 π  + 4 + 1 π  HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review Case 2. Roots of the Denominator of πΉ(π ) are Real and Repeated πΉ π  = 2 (π  + 1)(π  + 2)2 = πΎ1 π  + 1 + πΎ2 (π  + 2)2 + πΎ3 π  + 2 lim π ββ1 [(2.23) Γ (π  + 1)] βΉ lim π ββ1 2 π  + 2 = lim π ββ1 πΎ1 + (π  + 1)πΎ2 π  + 2 βΉ πΎ1 = 2 lim π ββ2 [(2.23) Γ (π  + 2)] βΉ lim π ββ2 2 π  + 1 = lim π ββ2 (π  + 2)πΎ1 π  + 1 + πΎ2 βΉ πΎ2 = β2 System Dynamics and Control 2.18 Modeling in Frequency Domain (2.22) (2.23) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
4. 4. 1/11/2016 4 Β§2.Laplace Transform Review πΉ π  = πΎ1 π  + 1 + πΎ2 (π  + 2)2 + πΎ3 π  + 2 πΎ1 = 2, πΎ2 = β2 (2.23) Γ (π  + 2)2 βΉ 2 π  + 1 = (π  + 2)2 πΎ1 π  + 1 + πΎ2 + (π  + 2)πΎ3 Differentiate (2.24) with respect to π  β2 π  + 1 2 = (π  + 2)πΎ1 π  + 1 2 + πΎ3 βΉ πΎ3 = β2 βΉ π π  = 2 π  + 1 β 2 π  + 2 2 β 2 π  + 2 Hence π¦ π‘ = 2πβπ‘ β 2π‘πβ2π‘ β 2πβ2π‘ (2.26) System Dynamics and Control 2.19 Modeling in Frequency Domain (2.23) (2.24) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 2.20 Modeling in Frequency Domain Β§2.Laplace Transform Review πΉ π  = 2 (π  + 1)(π  + 2)2 Matlab F=zpk([], [-1 -2 -2],2) Result F = 2 ------------------ (s+1) (s+2)^2 Continuous-time zero/pole/gain model (2.22) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 2.21 Modeling in Frequency Domain Β§2.Laplace Transform Review πΉ π  = 2 π 3 + 5π 2 + 8π  + 4 = 2 π  + 1 β 2 π  + 2 2 β 2 π  + 2 Matlab [r,p,k] = residue([2],[1,5,8,4]) Result r = [β2, β2, 2], p = [β2, β2, β1], k = [ ] πΉ π  = 0 π + (β2) π1 1 [π  β (β2 π1 )]2 + (β2 π2 ) 1 π  β(β2 π2 ) + 2 π3 1 π  β(β1 π3 ) = β 2 π  + 2 2 β 2 π  + 2 + 2 π  + 1 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review In general, given an πΉ(π ) whose denominator has real and distinct roots, a partial-fraction expansion πΉ π  = π(π ) π·(π ) = π(π ) π  + π1 π π  + π2 β¦ π  + ππ β¦ (π  + π π) = πΎ1 π  + π1 π + πΎ1 π  + π1 πβ1 + β― + πΎ2 π  + π1 + πΎ2 π  + π2 + β― + πΎπ π  + ππ + β― + πΎπ π  + π π β’ multiply (2.27) by π  + π1 π to get πΉ1 π  = π  + π1 π πΉ(π ) β’ let π  approach βππ System Dynamics and Control 2.22 Modeling in Frequency Domain (2.27) To find πΎπ πΎπ = 1 π β 1 ! π πβ1 πΉ1(π ) ππ  πβ1 π βπ1 π = 1,2, β¦, π; 0! = 1 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review Case 3. Roots of the Denominator of πΉ(π ) are Complex or Imaginary πΉ π  = 3 π (π 2 + 2π  + 5) = πΎ1 π  + πΎ2 π  + πΎ3 π 2 + 2π  + 5 lim π β0 [(2.31) Γ π ] βΉ πΎ1 = 3/5 First multiplying (2.31) by the lowest common denominator, π (π 2 + 2π  + 5), and clearing the fraction 3 = πΎ1 π 2 + 2π  + 5 + πΎ2 π  + πΎ3 π  (2.32) βΉ 3 = πΎ2 + 3 5 π 2 + πΎ3 + 6 5 π  + 3 Balancing the coefficients: πΎ2 = β3/5, πΎ3 = β6/5 πΉ π  = 3 π (π 2 + 2π  + 5) = 3 5 1 π  β 3 5 π  + 2 π 2 + 2π  + 5 System Dynamics and Control 2.23 Modeling in Frequency Domain (2.30) (2.33) (2.31) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien β π΄πβππ‘ πππ ππ‘ + π΅πβππ‘ π ππππ‘ = π΅ π +π +π΅π (π +π)2+π2 (Table 2.1 β 6&7) Β§2.Laplace Transform Review πΉ π  = 3 5 1 π  β 3 5 π  + 2 π 2 + 2π  + 5 = 3 5 1 π  β 3 5 π  + 1 + 1/2 2 (π  + 1)2+22 βΉ π π‘ = 3 5 β 3 5 πβπ‘ πππ 2π‘ + 1 2 π ππ2π‘ or π π‘ = 0.6 β 0.671πβπ‘ cos(2π‘ β π) (2.41) System Dynamics and Control 2.24 Modeling in Frequency Domain (2.38) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
5. 5. 1/11/2016 5 Β§2.Laplace Transform Review πΉ π  = 2 (π  + 1)(π  + 2)2 π¦ π‘ = 2πβπ‘ β 2π‘πβ2π‘ β 2πβ2π‘ (2.26) Matlab numf=2; denf=poly([-1 -2 -2]); [r,p,k]=residue(numf,denf) Result r = [-2 -2 2], p = [-2 -2 -1], k = [] πΉ π  = 0 π + β2 π1 1 [π  β (β2 π1 )]2 + (β2 π2 ) 1 π  β (β2 π2 ) + 2 π3 1 π  β (β1 π3 ) = β2 1 (π  + 2)2 β 2 1 π  + 2 + 2 1 π  + 1 System Dynamics and Control 2.25 Modeling in Frequency Domain (2.22) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review πΉ π  = 3 π (π 2 + 2π  + 5) Matlab F=tf([3],[1 2 5 0]) Result F = 3 ----------------------- s^3 + 2 s^2 + 5 s Continuous-time transfer function System Dynamics and Control 2.26 Modeling in Frequency Domain (2.30) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review In general, given an πΉ(π ) whose denominator has complex or purely imaginary roots, a partial-fraction expansion πΉ π  = π(π ) π·(π ) = π(π ) π  + π1 (π 2 + ππ  + π) β¦ = πΎ1 (π  + π1) + πΎ2 π  + πΎ3 (π 2 + ππ  + π) + β― To find πΎπ β’ the πΎπ βs in (2.42) are found through balancing the coefficients of the equation after clearing fractions β’ put (πΎ2 π  + πΎ3)/(π 2 + ππ  + π) in to the form π΅ π  + π + π΅π (π  + π)2+π2 System Dynamics and Control 2.27 Modeling in Frequency Domain (2.42) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review πΉ π  = 3 π (π 2 + 2π  + 5) π π‘ = 3 5 β 3 5 πβπ‘ πππ 2π‘ + 1 2 π ππ2π‘ Matlab syms s; f=ilaplace(3/(s*(s^2+2*s+5))); pretty(f) Result f = 3/5 - (3*exp(-t)*(cos(2*t) + sin(2*t)/2))/5 / sin(2 t) exp(-t) | cos(2 t) + ---------- | 3 3 2 / - - -------------------------------------- 5 5 System Dynamics and Control 2.28 Modeling in Frequency Domain (2.30) (2.38) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review πΉ π  = 3 π (π 2 + 2π  + 5) πΉ π  = 3/5 π  β 3 20 2+ π1 π  + 1 + π2 + 2 β π1 π  + 1 + π2 Matlab numf=3; denf=[1 2 5 0]; [r,p,k]=residue(numf,denf) Result r=[-0.3+0.15i; -0.3-0.15i; 0.6]; p=[-1+2i; -1-2i; 0]; k=[] πΉ π  = 0 π + (β0.3 + π0.15) π1 1 π  β (β1 + π2 π1 ) +(β0.3 β π0.15) π2 1 π  β (β1 β 2π π2 ) + (0.6) π3 1 π  β (0 π3 ) = β 0.3 β π0.15 π  + 1 β 2π β 0.3 + π0.15 π  + 1 + 2π + 0.6 1 π  System Dynamics and Control 2.29 Modeling in Frequency Domain (2.30) (2.47) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review Run ch2sp1 and ch2sp2 in Appendix F Learn how to use the Symbolic Math Toolbox to β’ construct symbolic objects β’ find the inverse Laplace transforms of frequency functions β’ find the Laplace of time functions System Dynamics and Control 2.30 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
6. 6. 1/11/2016 6 Β§2.Laplace Transform Review Skill-Assessment Ex.2.1 Problem Find the Laplace transform of π π‘ = π‘πβ5π‘ Solution πΉ π  = β π‘πβ5π‘ = 1 (π  + 5)2 Matlab syms t s F; f = t*exp(-5*t); F=laplace(f, s); pretty(F) Result 1 --------- 2 (s + 5) System Dynamics and Control 2.31 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review Skill-Assessment Ex.2.2 Problem Find the inverse Laplace transform of πΉ π  = 10 π (π  + 2)(π  + 3)2 Solution Expanding πΉ(π ) by partial fractions πΉ π  = π΄ π  + π΅ π  + 2 + πΆ (π  + 3)2 + π· π  + 3 π΄ = 10 (π  + 2)(π  + 3)2 π β0 = 5 9 , π΅ = 10 π (π  + 3)2 π ββ2 = β5 πΆ = 10 π (π  + 2) π ββ3 = 10 3 , π· = (π  + 3)2 ππΉ(π ) ππ  π ββ3 = 40 9 βΉ πΉ π  = 5 9 1 π  β 5 1 π  + 2 + 10 3 1 (π  + 3)2 + 40 9 1 π  + 3 System Dynamics and Control 2.32 Modeling in Frequency Domain where, HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§2.Laplace Transform Review πΉ π  = 5 9 1 π  β 5 1 π  + 2 + 10 3 1 (π  + 3)2 + 40 9 1 π  + 3 Taking the inverse Laplace transform π π‘ = 5 9 β 5πβ2π‘ + 10 3 π‘πβ3π‘ + 40 9 πβ3π‘ Matlab syms s; f=ilaplace(10/(s*(s+2)*(s+3)^2)); pretty(f) Result exp(-3 t) 40 t exp(-3 t) 10 5 --------------- - exp(-2 t) 5 + ---------------- + - 9 3 9 System Dynamics and Control 2.33 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 πΊ π  πΆ(π ) π π π π π(π‘) ππ‘ π + π πβ1 π πβ1 π π‘ ππ‘ πβ1 + β― + π0 π π‘ = π π π π π(π‘) ππ‘ π + π πβ1 π πβ1 π(π‘) ππ‘ πβ1 + β― + π0 π π‘ π(π‘): output π(π‘): input ππβs, ππβs: constant System Dynamics and Control 2.34 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function π π π π π(π‘) ππ‘ π + π πβ1 π πβ1 π π‘ ππ‘ πβ1 + β― + π0 π π‘ = π π π π π(π‘) ππ‘ π + π πβ1 π πβ1 π(π‘) ππ‘ πβ1 + β― + π0 π π‘ - Taking the Laplace transform of both sides with zero initial conditions π π π  π πΆ π  + π πβ1 π  πβ1 πΆ π  + β― + π0 πΆ π  = π π π  π π π  + π πβ1 π  πβ1 π π  + β― + π0 π π  - The transfer function πΊ π  = πΆ(π ) π(π ) = π π π  π + π πβ1 π  πβ1 + β― + π0 π π π  π + π πβ1 π  πβ1 + β― + π0 - The output of the system can be written in the form πΆ π  = πΊ π  π(π ) (2.54) System Dynamics and Control 2.35 Modeling in Frequency Domain (2.53) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function - Ex.2.4 Transfer Function for a Differential Equation Find the transfer function represented by ππ(π‘) ππ‘ + 2π(π‘) = π(π‘) Solution Taking the Laplace transform with zero initial conditions π πΆ π  + 2πΆ(π ) = π(π ) The transfer function πΊ π  = πΆ(π ) π(π ) = 1 π  + 2 System Dynamics and Control 2.36 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
7. 7. 1/11/2016 7 Β§3.The Transfer Function Run ch2p9 through ch2p12 in Appendix B Learn how to use MATLAB to β’ create transfer functions with numerators and denominators in polynomial or factored form β’ convert between polynomial and factored forms β’ plot time functions System Dynamics and Control 2.37 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function Run ch2sp3 in Appendix F Learn how to use the Symbolic Math Toolbox to β’ simplify the input of complicated transfer functions as well as improve readability β’ enter a symbolic transfer function and convert it to a linear time-invariant (LTI) object as presented in Appendix B, ch2p9 System Dynamics and Control 2.38 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function - Ex.2.5 System Response from the Transfer Function Given πΊ π  = 1/(π  + 2), find the response, π(π‘) to an input, π π‘ = π’(π‘), a unit step, assuming zero initial conditions Solution For a unit step π π‘ = π’ π‘ βΉ π π  = 1/π  The output πΆ π  = π π  πΊ π  = 1 π  1 π  + 2 Expanding by partial fractions πΆ π‘ = 1 2 1 π  β 1 2 1 π  + 2 Taking the inverse Laplace transform π π‘ = 0.5 β 0.5πβ2π‘ (2.60) System Dynamics and Control 2.39 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function πΊ π  = 1 π  + 2 π π  = 1 π  π π‘ = 1 2 β 1 2 πβ2π‘ Matlab syms s C=1/(s*(s+2)) C=ilaplace(C) Result C = 1/2 - exp(-2*t)/2 π π‘ = 1 2 β 1 2 πβ2π‘ System Dynamics and Control 2.40 Modeling in Frequency Domain (2.60) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function π π‘ = 1 2 β 1 2 πβ2π‘ Matlab t=0:0.01:1; plot(t,(1/2-1/2*exp(-2*t))) Result System Dynamics and Control 2.41 Modeling in Frequency Domain (2.60) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function Skill-Assessment Ex.2.3 Problem Find the transfer function, πΊ π  = πΆ(π )/π(π ), corresponding to the differential equation π3 π ππ‘3 + 3 π2 π ππ‘2 + 7 ππ ππ‘ + 5π = π2 π ππ‘2 + 4 ππ ππ‘ + 3π Solution Taking the Laplace transform with zero initial conditions π 3 πΆ π  + 3π 2 πΆ π  + 7π πΆ π  + 5πΆ π  = π 2 π π  + 4π π π  + 3π π  Collecting terms π 3 + 3π 2 + 7π  + 5 πΆ π  = π 2 + 4π  + 3 π(π ) The transfer function πΊ π  = πΆ(π ) π(π ) = π 2 + 4π  + 3 π 3 + 3π 2 + 7π  + 5 System Dynamics and Control 2.42 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
8. 8. 1/11/2016 8 Β§3.The Transfer Function Skill-Assessment Ex.2.4 Problem Find the differential equation corresponding to the transfer function πΊ π  = 2π  + 1 π 2 + 6π  + 2 Solution The transfer function πΊ π  = πΆ(π ) π(π ) = 2π  + 1 π 2 + 6π  + 2 Cross multiplying π2 π ππ‘2 + 6 ππ ππ‘ + 2π = 2 ππ ππ‘ + π System Dynamics and Control 2.43 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function Skill-Assessment Ex.2.5 Problem Find the ramp response for a system whose transfer function πΊ π  = π  (π  + 4)(π  + 8) Solution For a ramp response π π‘ = π‘π’ π‘ βΉ π π  = 1 π 2 The output πΆ π  = π π  πΊ π  = 1 π 2 π  (π  + 4)(π  + 8) = π΄ π  + π΅ π  + 4 + πΆ π  + 8 System Dynamics and Control 2.44 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§3.The Transfer Function πΆ π  = 1 π (π  + 4)(π  + 8) = π΄ π  + π΅ π  + 4 + πΆ π  + 8 π΄ = 1 (π  + 4)(π  + 8) π β0 = 1 32 π΅ = 1 π (π  + 8) π ββ4 = β 1 16 πΆ = 1 π (π  + 4) π ββ8 = 1 32 βΉ πΆ π  = 1 32 1 π  β 1 16 1 π  + 4 + 1 32 1 π  + 8 The ramp response π π‘ = 1 32 β 1 16 πβ4π‘ + 1 32 πβ8π‘ System Dynamics and Control 2.45 Modeling in Frequency Domain where, HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Summarizes the components and the relationships between voltage and current and between voltage and charge under zero initial conditions System Dynamics and Control 2.46 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 ππΆ(π )/π(π ) Solution The voltage loop πΏ ππ ππ‘ + ππ + 1 πΆ 0 1 π π ππ = π£(π‘) Using the relationships π π‘ = ππ(π‘)/ππ‘ and π = πΆπ£ πΆ πΏ π2 π ππ‘2 + π ππ ππ‘ + 1 πΆ π = π£(π‘) βΉ πΏπΆ π2 π£ πΆ ππ‘2 + ππΆ ππ£ πΆ ππ‘ + π£ πΆ = π£(π‘) System Dynamics and Control 2.47 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions πΏπΆ π2 π£ πΆ ππ‘2 + ππΆ ππ£ πΆ ππ‘ + π£ πΆ = π£(π‘) Taking Laplace transform assuming zero initial conditions πΏπΆπ 2 + ππΆπ  + 1 ππΆ π  = π(π ) Solving for the transfer function ππΆ(π ) π(π ) = 1 πΏπΆ π 2 + π πΏ π  + 1 πΏπΆ Block diagram of series RLC electrical network System Dynamics and Control 2.48 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
9. 9. 1/11/2016 9 Β§4.Electrical Network Transfer Functions Impedance - A resistance resists or βimpedesβ the flow of current. The corresponding relation is π£/π = π . 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 π(π ) to a current transform πΌ(π ) and thus implies a current source - Standard symbol for impedance π(π ) β‘ π(π ) πΌ(π ) - Kirchhoffβs voltage law to the transformed circuit [ Sum of Impedances ] Γ πΌ π  = [ Sum ofApplied Voltages ] (2.72) System Dynamics and Control 2.49 Modeling in Frequency Domain (2.70) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - The impedance of a resistor is its resistance π π  = π - For a capacitor π£ π‘ = 1 πΆ 0 π‘ πππ‘ βΉ π π  = πΌ(π ) πΆ π  π  The impedance of a capacitor π π  = 1 πΆπ  - For an inductor π£ π‘ = πΏ ππ ππ‘ βΉ π π  = πΏπΌ π  π  The impedance of a inductor π π  = πΏπ  System Dynamics and Control 2.50 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 System Dynamics and Control 2.51 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Series Impedances β’ Two impedances are in series if they have the same current. If so, the total impedance is the sum of the individual impedances π π£ π πΆ π π  = π1 π  + π2(π ) β’ Example: a resistor π and capacitor πΆ in series have the equivalent impedance π π  = π + 1 πΆπ  = ππΆπ  + 1 πΆπ  βΉ π(π ) πΌ(π ) β‘ π π  = ππΆπ  + 1 πΆπ  and the differential equation model is πΆ ππ£ ππ‘ = ππΆ ππ ππ‘ + π(π‘) System Dynamics and Control 2.52 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Parallel Impedances β’ Two impedances are in parallel if they have the same voltage difference across them. Their impedances combine by the reciprocal rule 1 π(π ) = 1 π1(π ) + 1 π2(π ) β’ Example: a resistor π and capacitor πΆ in parallel have the equivalent impedance 1 π(π ) = 1 1/πΆπ  + 1 π βΉ π(π ) πΌ(π ) β‘ π π  = π ππΆπ  + 1 and the differential equation model is ππΆ ππ£ ππ‘ + π£ = ππ(π‘) System Dynamics and Control 2.53 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Admittance π π  β‘ 1 π(π ) = πΌ(π ) π(π ) In general, admittance is complex β’ The real part of admittance is called conductance πΊ = 1 π β’ 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 System Dynamics and Control 2.54 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
10. 10. 1/11/2016 10 Β§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 Kirchhoffβs voltage law to the transformed circuit The steps are as follows 1.Redraw the original network showing all time variables, such as π£(π‘), π(π‘), and π£ πΆ(π‘), as Laplace transforms π(π ), πΌ(π ), and ππΆ(π ), 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 System Dynamics and Control 2.55 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Ex.2.7 Transfer Function - Single Loop via Transform Method Find the transfer function ππΆ(π )/π(π ) Solution The mess equation using impedances πΏπ  + π + 1 πΆπ  πΌ π  = π(π ) βΉ πΌ(π ) π(π ) = 1 πΏπ  + π + 1 πΆπ  The voltage across the capacitor ππΆ π  = πΌ(π ) 1 πΆπ  βΉ ππΆ(π )/π(π ) System Dynamics and Control 2.56 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Simple Circuits via Nodal Analysis - Ex.2.8 Transfer Function - Single Node via Transform Method Find the transfer function ππΆ(π )/π(π ) Solution The transfer function can be obtained by summing currents flowing out of the node whose voltage is ππΆ(π ) ππΆ(π ) 1/πΆπ  + ππΆ π  β π(π ) π + πΏπ  = 0 : the current flowing out of the node through the capacitor : the current flowing out of the node through the series resistor and inductor System Dynamics and Control 2.57 Modeling in Frequency Domain βΉ ππΆ(π ) π(π ) ππΆ(π ) πΌ/πΆπ  ππΆ π  β π(π ) π + πΏπ  HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Complex Circuits via Mesh Analysis To solve complex electrical networks - those with multiple loops and nodes β 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 Kirchhoffβs voltage law around each mesh 5.Solve the simultaneous equations for the output 6.Form the transfer function System Dynamics and Control 2.58 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Ex.2.10 Transfer Function β Multiple Loops Find the transfer function πΌ2(π )/π(π ) Solution Convert the network into Laplace transforms Summing voltages around each mesh through which the assumed currents flow π1 πΌ1 + πΏπ πΌ1 β πΏπ πΌ2 = π πΏπ πΌ2 + π2 πΌ2 + 1 πΆπ  πΌ2 β πΏπ πΌ1 = 0 π1 + πΏπ  πΌ1 β πΏπ πΌ2 = π βπΏπ πΌ1 + πΏπ  + π2 + 1 πΆπ  πΌ2 = 0 System Dynamics and Control 2.59 Modeling in Frequency Domain or (2.80) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Cramerβs Rule Consider a system of π linear equations for π unknowns, represented in matrix multiplication form as follows π΄π₯ = π π΄ : (π Γ π) matrix has a nonzero determinant π₯ : the column vector of the variables π₯ = (π₯1, β¦, π₯ π) π π : the column vector of known parameters The system has a unique solution, whose individual values for the unknowns are given by π₯π = det(π΄π) det(π΄) , π = 1, β¦ , π π΄π : the matrix formed by replacing the πth column of π΄ by the column vector π System Dynamics and Control 2.60 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
11. 11. 1/11/2016 11 Β§4.Electrical Network Transfer Functions π1 + πΏπ  πΌ1 β πΏπ πΌ2 = π βπΏπ πΌ1 + πΏπ  + π2 + 1 πΆπ  πΌ2 = 0 Using Cramerβs rule πΌ2 = π1 + πΏπ  π βπΏπ  0 π1 + πΏπ  βπΏπ  βπΏπ  πΏπ  + π2 + 1 πΆπ  = 0 β πΏπ π π1 + πΏπ  πΏπ  + π2 + 1 πΆπ  β πΏ2 π 2 = πΏπΆπ 2 π π1 + π2 πΏπΆπ 2 + π1 π2 πΆ + πΏ π  + π1 System Dynamics and Control 2.61 Modeling in Frequency Domain (2.80) (2.81) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions πΌ2 = πΏπΆπ 2 π π1 + π2 πΏπΆπ 2 + π1 π2 πΆ + πΏ π  + π1 Forming the transfer function πΊ π  = πΌ2 π  π(π ) = πΏπΆπ 2 π1 + π2 πΏπΆπ 2 + π1 π2 πΆ + πΏ π  + π1 The network is now modeled as the transfer function of figure System Dynamics and Control 2.62 Modeling in Frequency Domain (2.82) (2.81) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Note π1 + πΏπ  πΌ1 β πΏπ πΌ2 = π βπΏπ πΌ1 + πΏπ  + π2 + 1 πΆπ  πΌ2 = 0 System Dynamics and Control 2.63 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Run ch2sp4 in Appendix F Learn how to use the Symbolic Math Toolbox to β’ solve simultaneous equations using Cramerβs rule β’ solve for the transfer function in Eq. (2.82) using Eq. (2.80) System Dynamics and Control 2.64 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Complex Circuits via Nodal Analysis - Ex.2.11 Transfer Function β Multiple Nodes Find the transfer function ππΆ(π )/π(π ) Solution Sum of currents flowing from the nodes marked ππΏ(π ) and ππΆ(π ) ππΏ β π π1 + ππΏ πΏπ  + ππΏ β ππΆ π2 = 0 ππΆ 1/πΆπ  + ππΆ β ππΏ π2 = 0 πΊ1 + πΊ2 + 1 πΏπ  ππΏ β πΊ2 ππΆ = ππΊ1 βπΊ2 ππΏ + πΊ2 + πΆπ  ππΆ = 0 System Dynamics and Control 2.65 Modeling in Frequency Domain or (2.86) (2.85) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions πΊ1 + πΊ2 + 1 πΏπ  ππΏ β πΊ2 ππΆ = ππΊ1 βπΊ2 ππΏ + πΊ2 + πΆπ  ππΆ = 0 Solving for the transfer function ππΆ(π ) π(π ) = πΊ1 πΊ2 πΆ π  πΊ1 + πΊ2 π 2 + πΊ1 πΊ2 πΏ + πΆ πΏπΆ π  + πΊ2 πΏπΆ Block diagram of the network System Dynamics and Control 2.66 Modeling in Frequency Domain (2.86) (2.87) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
12. 12. 1/11/2016 12 Β§4.Electrical Network Transfer Functions - Another way to write node equations is to replace voltage sources by current sources. In order to handle multiple-node electrical networks, we can perform the following steps 1.Replace passive element values with their admittances 2.Replace all sources and time variables with their Laplace transform 3.Replace transformed voltage sources with transformed current sources 4.Write Kirchhoffβs current law at each node 5.Solve the simultaneous equations for the output 6.Form the transfer function System Dynamics and Control 2.67 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Norton's Theorem Any collection of batteries and resistances with two terminals is electrically equivalent to an ideal current source π in parallel with a single resistor π. The value of π is the same as that in the Thevenin equivalent and the current π can be found by dividing the open circuit voltage by π Β§4.Electrical Network Transfer Functions - Ex.2.12 Transfer Function β Multiple Nodes with Current Sources Find the transfer function ππΆ(π )/π(π ) Solution Convert all impedances to admittances and all voltage sources in series with an impedance to current sources in parallel with an admittance using Nortonβs theorem System Dynamics and Control 2.68 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Using the general relationship πΌ π  = π π  π(π ) and summing currents at the node ππΏ(π ) πΊ1 ππΏ π  + 1 πΏπ  ππΏ π  + πΊ2 ππΏ π  β ππΆ π  = πΊ1 π(π ) Summing the currents at the node ππΆ(π ) πΆππΆ π  + πΊ2 ππΆ π  β ππΏ π  = 0 (2.89) Solving (2.88) and (2.89), forming the transfer function ππΆ(π ) π(π ) = πΊ1 πΊ2 πΆ π  πΊ1 + πΊ2 π 2 + πΊ1 πΊ2 πΏ + πΆ πΏπΆ π  + πΊ2 πΏπΆ System Dynamics and Control 2.69 Modeling in Frequency Domain (2.88) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Note πΊ1 ππΏ π  + 1 πΏπ  ππΏ π  + πΊ2 ππΏ π  β ππΆ π  = πΊ1 π(π ) πΆππΆ π  + πΊ2 ππΆ π  β ππΏ π  = 0 (2.89) System Dynamics and Control 2.70 Modeling in Frequency Domain (2.88) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions A Problem-Solving Technique In all of the previous examples, we have seen a repeating pattern in the equations that we can use to our advantage. If we recognize this pattern, we need not write the equations component by component; we can sum impedances around a mesh in the case of mesh equations or sum admittances at a node in the case of node equations System Dynamics and Control 2.71 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Ex.2.13 Mesh Equations via Inspection Write the mesh equations for the network Solution The mesh equations for loop 1 + 2π  + 2 πΌ1 β 2π  + 1 πΌ2 β πΌ3 = π (2.94.a) System Dynamics and Control 2.72 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
13. 13. 1/11/2016 13 Β§4.Electrical Network Transfer Functions The mesh equations for loop 2 β 2π  + 2 πΌ1 + 9π  + 1 πΌ2 β 4π πΌ3 = 0 (2.94.b) System Dynamics and Control 2.73 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions The mesh equations for loop 3 βπΌ1 β 4π πΌ2 + 4π  + 1 + 1 π  πΌ3 = 0 (2.94.c) System Dynamics and Control 2.74 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions + 2π  + 2 πΌ1 β 2π  + 1 πΌ2 β πΌ3 = π (2.94.a) β 2π  + 2 πΌ1 + 9π  + 1 πΌ2 β 4π πΌ3 = 0 (2.94.b) βπΌ1 β 4π πΌ2 + 4π  + 1 + 1 π  πΌ3 = 0 Matlab syms s I1 I2 I3 V; A=[(2*s+2) -(2*s+1) -1; -(2*s+1) (9*s+1) -4*s; -1 -4*s (4*s+1+1/s)]; B=[I1;I2;I3]; C=[V;0;0]; B=inv(A)*C; pretty(B) System Dynamics and Control 2.75 Modeling in Frequency Domain (2.94.c) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Result / 3 2 | V (20 s + 13 s + 10 s + 1) | | ----------------------------------- | | #1 | | 3 2 | | V (8 s + 10 s + 3 s + 1) | | -------------------------------- | | #1 | | 2 | | V s (8 s + 13 s + 1) | | -------------------------- | #1 / where 4 3 2 #1 == 24 s + 30 s + 17 s + 16 s + 1 System Dynamics and Control 2.76 Modeling in Frequency Domain 20π 3 + 13π 2 + 10π  + 1 π 24π 4 + 30π 3 + 17π 2 + 16π  + 1 8π 3 + 10π 2 + 3π  + 1 π 24π 4 + 30π 3 + 17π 2 + 16π  + 1 π  8π 2 + 13π  + 1 π 24π 4 + 30π 3 + 17π 2 + 16π  + 1 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Operational Amplifiers - An operational amplifier (op-amp) is an electronic amplifier used as a basic building block to implement transfer functions - Op-amp has the following characteristics 1.Differential input, π£2 π‘ β π£1 π‘ 2.High input impedance, ππ = β (ideal) 3.Low output impedance, π π = 0 (ideal) 4.High constant gain amplification, π΄ = β (ideal) - The output, π£ π(π‘), is given by π£ π π‘ = π΄[π£2 π‘ β π£1 π‘ ] (2.95) System Dynamics and Control 2.77 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Inverting Operational Amplifiers - If π£2(π‘) is grounded, the amplifier is called an inverting operational amplifier - The output, π£ π(π‘), is given by π£ π π‘ = βπ΄π£1 π‘ (2.96) System Dynamics and Control 2.78 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
14. 14. 1/11/2016 14 Β§4.Electrical Network Transfer Functions Inverting Operational Amplifiers ππ π  = β β πΌ π(π ) = 0 πΌ1 π  = βπΌ2 π  π΄ = β β π£1 π‘ β 0 πΌ1 π  = ππ π  π1 π  = βπΌ2 π  = β ππ π  π2 π  The transfer function of the inverting operational amplifier ππ(π ) ππ(π ) = β π2(π ) π1(π ) System Dynamics and Control 2.79 Modeling in Frequency Domain (2.97) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Ex.2.14 Transfer Function β Inverting Op-Amp Circuit Find the transfer function ππ(π )/ππ(π ) Solution The impedances 1 π1 = 1 1/πΆ1 π  + 1 π1 β π1 = 1 πΆ1 π  + 1 π1 = 360 Γ 103 2.016π  + 1 π2 = π2 + 1 πΆ2 π  = 220 Γ 103 + 107 π  The transfer function ππ(π ) ππ(π ) = β π2 π  π1 π  = β 360 Γ 103 2.016π  + 1 220 Γ 103 + 107 π  = β1.232 π 2 + 45.95π  + 22.55 π  System Dynamics and Control 2.80 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Noninverting Operational Amplifiers ππ = π΄ ππ β π1 π1 = π1 π1 + π2 ππ The transfer function of the noninverting operational amplifier ππ(π ) ππ(π ) = π1 π  + π2(π ) π1(π ) System Dynamics and Control 2.81 Modeling in Frequency Domain (2.104) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions - Ex.2.15 Transfer Function β Noninverting Op-Amp Circuit Find the transfer function ππ(π )/ππ(π ) Solution The impedances π1 = π1 + 1 πΆ1 π  π2 = π2 1 πΆ2 π  π2 + 1 πΆ2 π  The transfer function ππ(π ) ππ(π ) = π1 π  + π2(π ) π1(π ) = πΆ2 πΆ1 π2 π1 π 2 + (πΆ2 π2 + πΆ1 π2 + πΆ1 π1)π  + 1 πΆ2 πΆ1 π2 π1 π 2 + πΆ2 π2 + πΆ1 π1 π  + 1 System Dynamics and Control 2.82 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Skill Assessment Ex.2.6 Problem Find πΊ π  = ππΏ(π )/π(π ) using mesh and nodal analysis Solution Mesh analysis Writing the mesh equations π  + 1 πΌ1 β π πΌ2 β πΌ3 = π βπ πΌ1 + 2π  + 1 πΌ2 β πΌ3 = 0 βπΌ1 β πΌ2 + π  + 2 πΌ3 = 0 System Dynamics and Control 2.83 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions π  + 1 πΌ1 β π πΌ2 β πΌ3 = π βπ πΌ1 + 2π  + 1 πΌ2 β πΌ3 = 0 βπΌ1 β πΌ2 + π  + 2 πΌ3 = 0 Solving the mesh equation for πΌ2 πΌ2 = π  + 1 π β1 βπ  0 β1 β1 0 π  + 2 π  + 1 βπ  β1 βπ  2π  + 1 β1 β1 β1 π  + 2 = π 2 + 2π  + 1 π π (π 2 + 5π  + 2) The voltage across πΏ ππΏ = π πΌ2 = π 2 + 2π  + 1 π π 2 + 5π  + 2 βΉ πΊ π  = ππΏ π = π 2 + 2π  + 1 π 2 + 5π  + 2 System Dynamics and Control 2.84 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
15. 15. 1/11/2016 15 Β§4.Electrical Network Transfer Functions Nodal analysis Writing the nodal equations 1 π  + 2 π1 β ππΏ = π βπ1 + 2 π  + 1 ππΏ = 1 π  π System Dynamics and Control 2.85 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions 1 π  + 2 π1 β ππΏ = π βπ1 + 2 π  + 1 ππΏ = 1 π  π Solving the nodal equation for ππΏ ππΏ = 1 π  + 2 π β1 1 π  1 π  + 2 β1 β1 2 π  + 1 = π 2 + 2π  + 1 π π 2 + 5π  + 2 βΉ πΊ π  = ππΏ π = π 2 + 2π  + 1 π 2 + 5π  + 2 System Dynamics and Control 2.86 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions Skill Assessment Ex.2.7 Problem If π1(π )is the impedance of a 10ππΉ capacitor and π2(π ) is the impedance of a 100πΞ© resistor, find the transfer function, πΊ π  = ππ(π )/ππ(π ) if these components are used with (a) an inverting op-amp and (b) a noninverting op-amp Solution π1 = π πΆ = 1 πΆπ  = 1 10β5 π  = 105 π  π2 = π π = π = 105 System Dynamics and Control 2.87 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§4.Electrical Network Transfer Functions (a) Inverting Op-Amp πΊ π  = β π2 π1 = β 105 105 π  = βπ  (b) Noninverting Op-Amp πΊ π  = π1 + π2 π1 = 105 π  + 105 105 π  = π  + 1 System Dynamics and Control 2.88 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions System Dynamics and Control 2.89 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions - Ex.2.16 Transfer Function - One Equation of Motion Find the transfer function π(π )/πΉ(π ) Solution Free body diagram Using Newtonβs law to sum all of the forces π π2 π₯(π‘) ππ‘2 + ππ£ ππ₯(π‘) ππ‘ + πΎπ₯ π‘ = π(π‘) Taking Laplace transform ππ 2 π π  + ππ£ π π π  + πΎπ π  = πΉ(π ) The transfer function πΊ(π ) = π(π ) πΉ(π ) = πΉ π  ππ 2 + ππ£ π  + πΎ System Dynamics and Control 2.90 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
16. 16. 1/11/2016 16 Β§5.Translational Mechanical System Transfer Functions Impedance - Define impedance for mechanical components π π π  β‘ πΉ π  π π  βΉ πΉ π  = π π π  π π  Sum of Impedances Γ π(π ) = Sum of Applied Forces - The impedance of a spring is its stiffness coefficient πΉ π  = πΎπ π  βΉ π π π  = πΎ (2.112) - For the viscous damper πΉ π  = ππ£ π π π  βΉ π π π  = ππ£ π  (2.113) - For the mass πΉ π  = ππ 2 π π  βΉ π π π  = ππ 2 (2.114) System Dynamics and Control 2.91 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions - Ex.2.17 Transfer Function - Two Degrees of Freedom Find the transfer function π2(π )/πΉ(π ) Solution Free body diagram of π1 The Laplace transform of the equation of motion of π1 + π1 π 2 + ππ£1 + ππ£3 π  + πΎ1 + πΎ2 π1 β ππ£3 π  + πΎ2 π2 = πΉ System Dynamics and Control 2.92 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions Free body diagram of π2 The Laplace transform of the equation of motion of π2 β ππ£3 π  + πΎ2 π1 + π2 π 2 + ππ£2 + ππ£3 π  + πΎ2 + πΎ3 π2 = 0 System Dynamics and Control 2.93 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions The Laplace transform of the equations of motion + π1 π 2 + ππ£1 + ππ£3 π  + πΎ1 + πΎ2 π1 β ππ£3 π  + πΎ2 π2 = πΉ β ππ£3 π  + πΎ2 π1 + π2 π 2 + ππ£2 + ππ£3 π  + πΎ2 + πΎ3 π2 = 0 The Laplace transform of the equations of motion π2 = π1 π 2 + ππ£1 + ππ£3 π  + πΎ1 + πΎ2 πΉ β ππ£3 π  + πΎ2 0 β = ππ£3 π  + πΎ2 πΉ β where β = π1 π 2 + ππ£1 +ππ£3 π + πΎ1 +πΎ2 β ππ£3 π  + πΎ2 β ππ£3 π  + πΎ2 π2 π 2 + ππ£2 +ππ£3 π + πΎ2 +πΎ3 The transfer function πΊ π  = π2(π ) πΉ(π ) = ππ£3 π  + πΎ2 β System Dynamics and Control 2.94 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions Note The Laplace transform of the equations of motion of π1 + π1 π 2 + ππ£1 + ππ£3 π  + πΎ1 + πΎ2 π1 β ππ£3 π  + πΎ2 π2 = πΉ The Laplace transform of the equations of motion of π2 β ππ£3 π  + πΎ2 π1 + π2 π 2 + ππ£2 + ππ£3 π  + πΎ2 + πΎ3 π2 = 0 System Dynamics and Control 2.95 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 π1 + π1 π 2 + ππ£1 + ππ£3 π  + πΎ1 + πΎ2 π1 β πΎ2 π2 β ππ£3 π3 = 0 System Dynamics and Control 2.96 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
17. 17. 1/11/2016 17 Β§5.Translational Mechanical System Transfer Functions The Laplace transform of the equations of motion of π2 βπΎ2 π1 + [π2 π 2 + ππ£2 + ππ£4 π  + πΎ2]π2 β ππ£4 π π3 = πΉ System Dynamics and Control 2.97 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions The Laplace transform of the equations of motion of π3 βππ£3 π π1 β ππ£4 π π2 + [π3 π 2 + ππ£3 + ππ£4 π ]π3 = 0 System Dynamics and Control 2.98 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions Skill-Assessment Ex.2.8 Problem Find the transfer function πΊ π  = π2(π ) πΉ(π ) Solution + π1 π 2 + ππ£1 + ππ£2 + ππ£3 π  + πΎ π1 β ππ£1 + ππ£2 + ππ£3 π  + πΎ π2 = πΉ βΉ +(π 2 + 3π  + 1)π1 β (3π  + 1)π2 = πΉ System Dynamics and Control 2.99 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions β ππ£1 + ππ£2 + ππ£3 π  + πΎ π1 + π2 π 2 + ππ£1 + ππ£2 + ππ£3 + ππ£4 π  + πΎ π2 = 0 βΉ β(3π  + 1)π1 + (π 2 + 4π  + 1)π2 = 0 System Dynamics and Control 2.100 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§5.Translational Mechanical System Transfer Functions + π 2 + 3π  + 1 π1 β 3π  + 1 π2 = πΉ β(3π  + 1)π1 + (π 2 + 4π  + 1)π2 = 0 The solution for π2 π2 = π 2 + 3π  + 1 πΉ β 3π  + 1 0 β = 3π  + 1 πΉ β where β= π 2 + 3π  + 1 β 3π  + 1 β 3π  + 1 π 2 + 4π  + 1 = π (π 3 + 7π 2 + 5π  + 1) βΉ πΊ π  = π2(π ) πΉ(π ) = 3π  + 1 π (π 3 + 7π 2 + 5π  + 1) System Dynamics and Control 2.101 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions System Dynamics and Control 2.102 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
18. 18. 1/11/2016 18 Β§6.Rotational Mechanical System Transfer Functions - Ex.2.19 Transfer Function β Two Equations of Motion Find the transfer function, π2(π )/π(π ), 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 Solution First, obtain the schematic from the physical system System Dynamics and Control 2.103 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions Next, draw a free-body diagram of π½1 vΓ  π½2, using superposition π½1 π 2 + π·1 π  + πΎ π1 π  β πΎπ2 π  = π(π ) (2.127.a) βπΎ1 π1 π  + (π½2 π 2 + π·2 π  + πΎ)π2 π  = 0 (2.127.b) System Dynamics and Control 2.104 Modeling in Frequency Domain Final free-body diagram for π½2 Torques on π½2 due only to the motion of π½2 Torques on π½2 due only to the motion of π½1 Final free-body diagram for π½1 Torques on π½1 due only to the motion of π½1 Torques on π½1 due only to the motion of π½2 HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions π½1 π 2 + π·1 π  + πΎ π1 π  β πΎπ2 π  = π(π )(2.127.a) βπΎ1 π1 π  + (π½2 π 2 + π·2 π  + πΎ)π2 π  = 0 (2.127.b) The solution for π2 π2 = π½1 π 2 + π·1 π  + πΎ π βπΎ 0 β = πΎπ β where β= π½1 π 2 + π·1 π  + πΎ βπΎ βπΎ π½2 π 2 + π·2 π  + πΎ βΉ πΊ π  = π2(π ) π(π ) = πΎ β System Dynamics and Control 2.105 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions Note π½1 π 2 + π·1 π  + πΎ π1 π  β πΎπ2 π  = π(π ) (2.127.a) βπΎ1 π1 π  + (π½2 π 2 + π·2 π  + πΎ)π2 π  = 0 (2.127.b) System Dynamics and Control 2.106 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 π½1 + π½1 π 2 + π·1 π  + πΎ π1 β πΎπ2 β 0π3 = π(π ) (2.131.a) System Dynamics and Control 2.107 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions The Laplace transform of the equations of motion of π½2 βπΎπ1 + π½2 π 2 + π·2 π  + πΎ π2 β π·2 π π3 = 0 (2.131.b) System Dynamics and Control 2.108 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
19. 19. 1/11/2016 19 Β§6.Rotational Mechanical System Transfer Functions The Laplace transform of the equations of motion of π½3 β0π1 β π·2 π π2 + π½3 π 2 + π·3 π  + π·2 π  π3 = 0 (2.131.c) System Dynamics and Control 2.109 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions Skill-Assessment Ex.2.9 Problem Find the transfer function πΊ π  = π2(π ) π(π ) Solution The equations of motion + π 2 + π  + 1 π1 π  β (π  + 1)π2 π  = π(π ) β π  + 1 π1 π  + (2π  + 2)π2 π  = 0 System Dynamics and Control 2.110 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§6.Rotational Mechanical System Transfer Functions The equations of motion π 2 + π  + 1 π1 π  β (π  + 1)π2 π  = π(π ) β π  + 1 π1 π  + (2π  + 2)π2 π  = 0 Solving for π2(π ) π2 = π 2 + π  + 1 π β π  + 1 0 π 2 + π  + 1 β(π  + 1) β π  + 1 2π  + 2 = (π  + 1)π 2π 3 + 3π 2 + 2π  + 1 βΉ πΊ π  = π2(π ) π(π ) = π  + 1 2π 3 + 3π 2 + 2π  + 1 System Dynamics and Control 2.111 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears Kinematic relationship π2 π1 = π1 π2 = π1 π2 Power on gears π1 π1 = π2 π2 The ratio of torques on two gears π2 π1 = π1 π2 = π2 π1 π1, π2 : rotation angles of gear 1 and 2, πππ π1, π2 : radius of gear 1 and 2, π π1, π2 : number of teeth of gear 1 and 2 π1, π2 : torques on gear 1 and 2, ππ System Dynamics and Control 2.112 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with 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 π1 without the gears Can the mechanical impedances be reflected from the output to the input, thereby eliminating the gears? System Dynamics and Control 2.113 Modeling in Frequency Domain b.equivalentsystem at the output after reflection of input torque a.rotational system driven by gears HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears π1 can be reflected to the output by multiplying by π2/π1 π½π 2 + π·π  + πΎ π2 π  = π1(π ) π2 π1 Convert π2(π ) into an equivalent π1(π ), so that π½π 2 + π·π  + πΎ π1 π2 π1 π  = π1(π ) π2 π1 System Dynamics and Control 2.114 Modeling in Frequency Domain b.equivalentsystem at the output after reflection of input torque a.rotational system driven by gears (2.131) (2.132) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
20. 20. 1/11/2016 20 Β§7.Transfer Functions for Systems with Gears π½π 2 + π·π  + πΎ π2 π  = π1(π )(π2/π1) (2.131) π½π 2 + π·π  + πΎ (π2/π1)π1 π  = π1(π )(π2/π1) (2.132) βΉ π½ π1 π2 2 π 2 + π· π1 π2 2 π  + πΎ π1 π2 2 π1 π  = π1(π ) Thus, the load can be thought of as having been reflected from the output to the input System Dynamics and Control 2.115 Modeling in Frequency Domain (2.133) b.equivalentsystem at the output after reflection of input torque c.equivalent system at the input after reflection of impedances a.rotational system driven by gears HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 where the impedance to be reflected is attached to the source shaft and is being reflected to the destination shaft System Dynamics and Control 2.116 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears - Ex.2.21 Transfer Function - System with Lossless Gears Find the transfer function, π2(π )/π1(π ), for the system Solution Reflect the impedances (π½1 and π·1) and torque (π1) on the input shaft to the output, where the impedances are reflected by (π1/π2)2 and the torque is reflected by (π1/π2) The equation of motion can now be written as π½π π 2 + π·π π  + πΎπ π2 π  = π1 π  (π2/π1) (2.139) System Dynamics and Control 2.117 Modeling in Frequency Domain b.system after reflection of torques and impedances to the output shaft a.rotationalmechanicalsystemwithgears HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears π½π π 2 + π·π π  + πΎπ π2 π  = π1 π  (π2/π1) (2.139) where, π½ π = π½1(π2/π1)2 +π½2, π·π = π·1(π2/π1)2 +π·2, πΎπ = πΎ2 Solving for πΊ(π ) πΊ π  = π2(π ) π1(π ) = π2/π1 π½ π π 2 + π·π π  + πΎπ System Dynamics and Control 2.118 Modeling in Frequency Domain b.system after reflection of torques and impedances to the output shaft a.rotationalmechanicalsystemwithgears HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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. π4 = π1 π3 π5 π2 π4 π6 π1 - For gear trains, the equivalent gear ratio is the product of the individual gear ratios System Dynamics and Control 2.119 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears - Ex.2.22 Transfer Function β Gears with Loss Find the transfer function, π1(π )/π1(π ), for the system Solution Reflect all of the impedances to the input shaft, π1 The equation of motion can now be written as π½π π 2 + π·π π  π1 π  = π1 π  The transfer function πΊ π  = π1 π  /π1 π  = 1/(π½ π π 2 + π·π π ) System Dynamics and Control 2.120 Modeling in Frequency Domain b.equivalent system at the inputa.system using a gear train HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
21. 21. 1/11/2016 21 Β§7.Transfer Functions for Systems with Gears Skill-Assessment Ex.2.10 Problem Find the TF πΊ π  = π2(π ) π(π ) Solution Transforming the network to one without gears by reflecting the 4ππ/πππ spring to the left and multiplying by (25/50)2 4[ππ/πππ] Γ 25 50 2 = 1[ππ/πππ] System Dynamics and Control 2.121 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears The equation of motion + π 2 + π  π1 π  β π π π π  = π(π ) βπ π1 π  + (π  + 1)π π π  = 0 System Dynamics and Control 2.122 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§7.Transfer Functions for Systems with Gears The equation of motion π 2 + π  π1 π  β π π π π  = π(π ) βπ π1 π  + (π  + 1)π π π  = 0 Solving for π π(π ) π π π  = π 2 + π  π βπ  0 π 2 + π  βπ  βπ  π  + 1 = π π(π ) π 3 + π 2 + π  βΉ π π π  π(π ) = 1 π 2 + π  + 1 The transfer function π2 π  π(π ) = 1 2 π π π  π(π ) = 1/2 π 2 + π  + 1 System Dynamics and Control 2.123 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions System Dynamics and Control 2.124 Modeling in Frequency Domain NASA flight simulator robot arm with electromechanical control system components HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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) β’ Fixed field: a magnetic field is developed by stationary permanent magnets or a stationary electromagnet β’ Armature: a rotating circuit, through which current π π(π‘) flows, passes through this magnetic field at right angles and feels a force πΉ = π΅ππ π(π‘) π΅ : the magnetic field strength π : the length of the conductor System Dynamics and Control 2.125 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions β’ A conductor moving at right angles to a magnetic field generates a voltage at the terminals of the conductor equal to π = π΅ππ£ π : the voltage π£ : the velocity of the conductor normal to the magnetic field β’ The current-carrying armature is rotating in a magnetic field, its voltage is proportional to speed π£ π π‘ = πΎπ π π(π‘) (2.144) π£ π π‘ : the back electromotive force (back emf) πΎπ : a constant of proportionality called the back emf constant π π(π‘): the angular velocity of the motor System Dynamics and Control 2.126 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
22. 22. 1/11/2016 22 Β§8.Electromechanical System Transfer Functions β’ Taking the Laplace transform ππ(π ) = πΎπ π π π(π ) (2.145) β’ The relationship between the armature current, π π(π‘), the applied armature voltage, π π(π‘), and the back emf, π£ π(π‘) π π πΌ π π  + πΏ π π πΌ π π  + ππ π  = πΈ π π  (2.146) β’ The torque developed by the motor is proportional to the armature current π π π  = πΎπ‘ πΌ π π  (2.147) π π : the torque developed by the motor πΎπ‘ : the motor torque constant, which depends on the motor and magnetic field characteristics System Dynamics and Control 2.127 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien ππ(π ) = πΎπ π π π(π ) (2.145), π π πΌ π π  + πΏ π π πΌ π π  + ππ π  = πΈ π π  (2.146) Β§8.Electromechanical System Transfer Functions β’ Rearranging Eq.(2.147) πΌ π(π ) = 1 πΎπ‘ π π(π ) β’ To find the TF of the motor, first substitute Eqs. (2.145) and (2.148) into (2.146), yielding (π π + πΏ π π )π π(π ) πΎπ‘ + πΎπ π π π π  = πΈ π π  β’ Then, find π π(π ) in terms of π π(π ), separate the input and output variables and obtain the transfer function, π π(π )/πΈ π(π ) System Dynamics and Control 2.128 Modeling in Frequency Domain (2.148) (2.149) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien (π π+πΏ π π )π π(π ) πΎπ‘ + πΎπ π π π π  = πΈ π π  (2.149) Β§8.Electromechanical System Transfer Functions β’ A typical equivalent mechanical loading on a motor π½ π : the equivalent inertia at the armature and includes both the armature inertia and, the load inertia reflected to the armature π· π : the equivalent viscous damping at the armature and includes both the armature viscous damping and, the load viscous damping reflected to the armature π π π  = (π½ π π 2 + π· π π )π π(π ) (2.150) β’ Substituting Eq.(2.150) into Eq.(2.149) (π π + πΏ π π )(π½ π π 2 + π· π π )π π(π ) πΎπ‘ + πΎπ π π π π  = πΈπ(π ) System Dynamics and Control 2.129 Modeling in Frequency Domain (2.151) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien (π π+πΏ π π )(π½ π π 2+π· π π )π π(π ) πΎπ‘ + πΎπ π π π π  = πΈπ(π ) (2.151) Β§8.Electromechanical System Transfer Functions β’ Assume that the armature inductance, πΏ π, is small compared to the armature resistance, π π, which is usual for a dc motor, Eq. (2.151) becomes π π πΎπ‘ π½ π π  + π· π + πΎπ π π π π  = πΈ π(π ) β’ After simplification π π(π ) πΈ π(π ) = πΎπ‘ π π 1 π½ π π  π  + 1 π½ π π· π + πΎπ‘ π π πΎπ β’ The form of Eq.(2.153) π π(π ) πΈ π(π ) = πΎ π (π  + πΌ) System Dynamics and Control 2.130 Modeling in Frequency Domain (2.153) (2.154) (2.152) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions β’ Let us first discuss the mechanical constants, π½ π and π· π. Consider the figure: a motor with inertia π½ π and damping π· π at the armature driving a load consisting of inertia π½ πΏ and damping π· πΏ Assuming that all inertia and damping values shown are known, π½ πΏ and π· πΏ can be reflected back to the armature as some equivalent inertia and damping to be added to π½ π and π· π , respectively. Thus, the equivalent inertia, π½ π , and equivalent damping, π· π, at the armature are π½ π = π½ π + π½πΏ π1 π2 2 π· π = π· π + π·πΏ π1 π2 2 System Dynamics and Control 2.131 Modeling in Frequency Domain (2.155.a) (2.155.b) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien ππ(π ) = πΎπ π π π(π ) (2.145), π π πΌπ π  + πΏ π π πΌπ π  + ππ π  = πΈπ π  (2.146), πΌπ(π ) = 1 πΎπ‘ π π(π ) (2.148) Β§8.Electromechanical System Transfer Functions β’ Substituting Eqs.(2.145), (2.148) into Eq. (2.146), with πΏ π = 0 π π πΎπ‘ π π π  + πΎπ π π π π  = πΈ π π  Taking the inverse Laplace transform π π πΎπ‘ π π π‘ + πΎπ π π π‘ = π π π‘ β’ When the motor is operating at steady state with a dc voltage input π π πΎπ‘ π π + πΎπ π π = π π βΉ π π = β πΎπ πΎπ‘ π π π π + πΎπ‘ π π π π System Dynamics and Control 2.132 Modeling in Frequency Domain (2.156) (2.157) (2.158) (2.159) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
23. 23. 1/11/2016 23 Β§8.Electromechanical System Transfer Functions The stall torque ππ π‘πππ = πΎπ‘ π π π π The no-load speed π ππβππππ = π π πΎπ The electrical constants of the motor πΎπ‘ π π = ππ π‘πππ π π πΎπ = π π π ππβππππ The electrical constants, πΎπ‘/π π and πΎπ, can be found from a dynamometer test of the motor, which would yield ππ π‘πππ and π ππβππππ for a given π π System Dynamics and Control 2.133 Modeling in Frequency Domain (2.160) (2.161) (2.162) (2.163) HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions - Ex.2.23 Transfer Function-DC Motor and Load Given the system and torque-speed curve, find the TF, System Dynamics and Control 2.134 Modeling in Frequency Domain π πΏ π  πΈ π π  HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions Solution Find the mechanical constants π½ π = π½ π + π½πΏ π1 π2 2 = 5 + 700 Γ 100 1000 2 = 12 π· π = π· π + π· πΏ π1 π2 2 = 2 + 800 Γ 100 1000 2 = 10 System Dynamics and Control 2.135 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions Find the electrical constants from the torque-speed curve ππ π‘πππ = 500, π ππβππππ = 50, π π = 100 πΎπ‘ π π = ππ π‘πππ π π = 500 100 = 5 πΎπ = π π π ππβππππ = 100 50 = 2 System Dynamics and Control 2.136 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions The transfer function π π(π )/πΈ π(π ) π π(π ) πΈ π(π ) = πΎπ‘ π π 1 π½ π π  π  + 1 π½ π π· π + πΎπ‘ π π πΎπ = 5 Γ 1 12 π  π  + 1 12 Γ 10 + 5 Γ 2 = 0.417 π (π  + 1.667) The transfer function π πΏ(π )/πΈ π(π ) π πΏ(π ) πΈ π(π ) = π π(π ) π1 π2 πΈ π(π ) = 0.417 Γ 100 1000 π (π  + 1.667) = 0.0417 π (π  + 1.667) System Dynamics and Control 2.137 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions Skill-Assessment Ex.2.11 Problem Find the TF, πΊ π  = π πΏ(π )/πΈπ (π ), for the motor and load system. The torque-speed curve is given by π π = β 8π π + 200 when the input voltage is 100π£πππ‘π  Solution Find the mechanical constants π½ π = π½ π + π½πΏ π1 π2 π3 π4 2 = 1 + 400 Γ 20 100 Γ 25 100 2 = 2 π· π = π· π + π·πΏ π1 π2 π3 π4 2 = 5 + 800 Γ 20 100 Γ 25 100 2 = 7 System Dynamics and Control 2.138 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
24. 24. 1/11/2016 24 π π = β8π π + 200 Β§8.Electromechanical System Transfer Functions Find the electrical constants from the torque-speed eq. π π = 0 βΉ π π = 200 π π = 0 βΉ π ππβππππ = 200/8 = 25 πΎπ‘ π π = ππ π‘πππ πΈ π = 200 100 = 2 πΎπ = πΈ π π ππβππππ = 100 25 = 4 System Dynamics and Control 2.139 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§8.Electromechanical System Transfer Functions Substituting all values into the motor transfer function π π(π ) πΈ π(π ) = πΎπ‘ π π 1 π½ π π  π  + 1 π½ π π· π + πΎπ‘ π π πΎπ = 2 Γ 1 2 π  π  + 1 2 7 + 2 Γ 4 = 1 π (π  + 7.5) The transfer function π πΏ(π )/πΈ π(π ) π πΏ(π ) πΈ π(π ) = π π(π ) π1 π2 π3 π4 πΈ π(π ) = 20 100 Γ 25 100 π (π  + 7.5) = 0.05 π (π  + 7.5) System Dynamics and Control 2.140 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 β’ When compared with mesh equations, the resulting electrical circuit is called a series analog β’ When compared with nodal equations, the resulting electrical circuit is called a parallel analog System Dynamics and Control 2.141 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§9.Electric Circuit Analogs Series Analog Consider the translational mechanical system, the equation of motion ππ 2 + ππ£ π  + πΎ π π  = πΉ π  = ππ 2 + ππ£ π  + πΎ π  π π π  βΉ ππ  + ππ£ + πΎ π  π π  = πΉ(π ) Kirchhoffβs mesh equation for the simple series RLC network πΏπ  + π + 1 πΆπ  πΌ π  = πΈ(π ) System Dynamics and Control 2.142 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§9.Electric Circuit Analogs Parameters for series analog mass = π βΉ inductor πΏ = π henries viscous damper = ππ£ βΉ resistor π = ππ£ohms spring = πΎ βΉ capacitor πΆ = 1/πΎ farads applied force = π(π‘) βΉ voltage source π π‘ = π(π‘) velocity = π£(π‘) βΉ mesh current π π‘ = π£(π‘) System Dynamics and Control 2.143 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§9.Electric Circuit Analogs - Ex.2.24 Converting a Mechanical System to a Series Analog Draw a series analog for the mechanical system Solution The equations of motion with π(π ) β π(π ) π1 π  + ππ£1 + ππ£3 + πΎ1 + πΎ2 π  π1 π  β ππ£3 + πΎ2 π  π2 π  = πΉ(π ) β ππ£3 + πΎ2 π  π1 π  + π2 π  + ππ£2 + ππ£3 + πΎ2 + πΎ3 π  π2 π  = 0 System Dynamics and Control 2.144 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
25. 25. 1/11/2016 25 + π1 π  + ππ£1 + ππ£3 + πΎ1+πΎ2 π  π1 π  β ππ£3 + πΎ2 π  π2 π  = πΉ(π ) β ππ£3 + πΎ2 π  π1 π  + π2 π  + ππ£2 + ππ£3 + πΎ2+πΎ3 π  π2 π  = 0 Β§9.Electric Circuit Analogs Coefficients represent sums of electrical impedance. Mechanical impedances associated with π1 form the first mesh, where impedances between the two masses are common to the two loops. Impedances associated with π2 form the second mesh π£1(π‘) and π£2(π‘)are the velocities of π1 and π2, respectively System Dynamics and Control 2.145 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§9.Electric Circuit Analogs Parallel Analog Consider the translational mechanical system, the equation of motion ππ  + ππ£ + πΎ π  π π  = πΉ(π ) Kirchhoffβs nodal equation for the simple parallel RLC network πΆπ  + 1 ππ  + 1 πΏπ  πΈ π  = πΌ(π ) System Dynamics and Control 2.146 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§9.Electric Circuit Analogs Parameters for parallel analog mass = π βΉ capacitor πΆ = π farads viscous damper = ππ£ βΉ resistor π = 1/ππ£ ohms spring = πΎ βΉ inductor πΏ = 1/πΎ henries applied force = π(π‘) βΉ current source π(π‘) = π(π‘) velocity = π£(π‘) βΉ node voltage π(π‘) = π£(π‘) System Dynamics and Control 2.147 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§9.Electric Circuit Analogs - Ex.2.25 Converting a Mechanical System to a Parallel Analog Draw a parallel analog for the mechanical system Solution The equations of motion with π(π ) β π(π ) π1 π  + ππ£1 + ππ£3 + πΎ1 + πΎ2 π  π1 π  β ππ£3 + πΎ2 π  π2 π  = πΉ(π ) β ππ£3 + πΎ2 π  π1 π  + π2 π  + ππ£2 + ππ£3 + πΎ2 + πΎ3 π  π2 π  = 0 System Dynamics and Control 2.148 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien + π1 π  + ππ£1 + ππ£3 + πΎ1+πΎ2 π  π1 π  β ππ£3 + πΎ2 π  π2 π  = πΉ(π ) β ππ£3 + πΎ2 π  π1 π  + π2 π  + ππ£2 + ππ£3 + πΎ2+πΎ3 π  π2 π  = 0 Β§9.Electric Circuit Analogs Coefficients represent sums of electrical admittances. Admittances associated with π1 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 π2 form the elements connected to the second node π£1(π‘) and π£2(π‘) are the velocities of π1 and π2, respectively System Dynamics and Control 2.149 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§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 + π½1 π 2 + π·1 π  + πΎ π1 π  β πΎπ2 π  = π(π ) βπΎπ1 π  + π½2 π 2 + π·2 π  + πΎ π2 π  = 0 System Dynamics and Control 2.150 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
26. 26. 1/11/2016 26 Β§9.Electric Circuit Analogs π½1 π 2 + π·1 π  + πΎ π1 π  β πΎπ2 π  = π(π ) βπΎπ1 π  + π½2 π 2 + π·2 π  + πΎ π2 π  = 0 Leting π1 π  = π1(π )/π , π2 π  = π2(π )/π  π½1 π  + π·1 + πΎ π  π1 π  β πΎ π  π2 π  = π(π ) β πΎ π  π1 π  + π½2 π  + π·2 + πΎ π  π2 π  = 0 From these equations, draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively System Dynamics and Control 2.151 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§10.Nonlinearities - A linear system possesses two properties β’ Superposition β’ Homogeneity - Some examples of physical nonlinearities System Dynamics and Control 2.152 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization - To obtain transfer function from a nonlinear system β’ Recognize the nonlinear component and write the nonlinear differential equation β’ Find the steady-state solution is called equilibrium β’ Linearize the nonlinear differential equation β’ Take the Laplace transform of the linearized differential equation, assuming zero initial conditions β’ Separate input and output variables and form the transfer function System Dynamics and Control 2.153 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization - Assume a nonlinear system operating at point π΄, [π₯0, π(π₯0)], small changes in the input can be related to changes in the output about the point by way of the slope of the curve at the point π΄ π π₯ β π(π₯0) β π π(π₯ β π₯0) βΉ πΏπ π₯ β π π πΏπ₯ βΉ π π₯ β π π₯0 + π π π₯ β π₯0 β π π₯0 + π π πΏπ₯ π π : the slope of the curve at point π΄ Ξ΄π₯ : small excursions of the input about point π΄ πΏπ(π₯): small changes in the output related by the slope at point π΄ System Dynamics and Control 2.154 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization - Ex.2.26 Linearizing a Function Linearize π π₯ = 5πππ π₯ about π₯ = π/2 Solution Using the linearized equation π π₯ β π π₯0 + π π πΏπ₯ where π π 2 = 5πππ  π 2 = 0 π π = ππ ππ₯ π₯= π 2 = (β5π πππ₯) π₯= π 2 = β5 The system can be presented as π π₯ β β5πΏπ₯ for small excursions of π₯ about π/2 System Dynamics and Control 2.155 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization Taylor series expansion Taylor series expansion expresses the value of a function in terms of the value of that function at a particular point, the excursion away from that point, and derivatives evaluated at that point π π₯ = π π₯0 + ππ ππ₯ π₯=π₯0 (π₯ β π₯ π) 1! + π2 π ππ₯2 π₯=π₯0 (π₯ β π₯ π)2 2! + β― For small excursions of π₯ from π₯0, the higher-order terms can be neglected π π₯ = π π₯0 + ππ ππ₯ π₯=π₯0 (π₯ β π₯ π) System Dynamics and Control 2.156 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
27. 27. 1/11/2016 27 Β§11.Linearization - Ex.2.27 Linearizing a Differential Equation Linearize the following equation for small excursion about π₯ = π/4 π2 π₯ ππ‘2 + 2 ππ₯ ππ‘ + πππ π₯ = 0 Solution The presence of the term πππ π₯ makes this equation nonlinear Since we want to linearize the equation about π₯ = π/4, we let π₯ = π/4 + πΏπ₯, where πΏπ₯ is the small excursion about π/4 π2 (πΏπ₯ + π/4) ππ‘2 + 2 π(πΏπ₯ + π/4) ππ‘ + πππ (πΏπ₯ + π/4) = 0 π2 (πΏπ₯ + π/4) ππ‘2 = π2 πΏπ₯ ππ‘2 π(πΏπ₯ + π/4) ππ‘ = ππΏπ₯ ππ‘ System Dynamics and Control 2.157 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization π π₯ = π π₯0 + ππ ππ₯ π₯=π₯0 (π₯ β π₯0) π π₯ = πππ π₯ = πππ  πΏπ₯ + π/4 π π₯0 = π π/4 = cos(π/4) = 2/2 π₯ β π₯0 = πΏπ₯ ππ ππ₯ π₯=π₯0 = ππππ π₯ ππ₯ π₯=π/4 = β sin π/4 = β 2/2 βΉ πππ  πΏπ₯ + π/4 = 2 2 + β 2 2 πΏπ₯ The linearized differential equation π2 πΏπ₯ ππ‘2 + 2 ππΏπ₯ ππ‘ β 2 2 πΏπ₯ = β 2 2 Solve this equation for πΏπ₯, and obtain π₯ = πΏπ₯ + π/4 System Dynamics and Control 2.158 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization - Ex.2.28 Transfer Function-Nonlinear Electrical Network Find the transfer function, ππΏ(π )/π(π ), for the electrical network, which contains a nonlinear resistor whose voltage-current relationship is defined by π π = 2π0.1π£ π , where π π and π£π are the resistor current and voltage, respectively. Also, π£(π‘) is a small-signal source Solution From the voltage-current relationship π π = 2π0.1π£ π βΉ π£π = 10ln(0.5π π) = 10ln(0.5π) Applying Kirchhoffβs voltage law around the loop πΏ ππ ππ‘ + 10 ln 0.5π β 20 = π£(π‘) System Dynamics and Control 2.159 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization - Evaluate the equilibrium solution β’ Set the small-signal source, π£(π‘), equal to zero β’ Evaluate the steady-state current In the steady state π£ πΏ π‘ = πΏππ/ππ‘ and ππ/ππ‘, given a constant battery source. Hence, the resistor voltage, π£π, is 20π π π = 2π0.1π£ π = 2π0.1Γ20 = 14.78π΄ βΉ π0 = π π = 14.78π΄ π0 is the equilibrium value of the network current βΉ π = π0 + πΏπ πΏ ππ ππ‘ + 10 ln 0.5π β 20 = π£(π‘) βΉ πΏ π(π0 + πΏπ) ππ‘ + 10ln[0.5 π0 + πΏπ ] β 20 = π£(π‘) System Dynamics and Control 2.160 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien πΏ π(π0+πΏπ) ππ‘ + 10 ln[0.5 π0 + πΏπ ] β 20 = π£(π‘) Β§11.Linearization π π = π π0 + ππ ππ π=π0 (π β π0) π π = ln(0.5π) = ln[0.5 π0 + πΏπ ] π π0 = ln(0.5π0) π β π0 = πΏπ ππ ππ π=π0 = π ln(0.5π) ππ π=π0 = 1 π π=π0 = 1 π0 βΉ ln[0.5 π0 + πΏπ ] = ln(0.5π0) + 1 π0 πΏπ The linearized equation πΏ ππΏπ ππ‘ + 10 ln(0.5π0) + 1 π0 πΏπ β 20 = π£(π‘) System Dynamics and Control 2.161 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization The linearized equation with πΏ = 1π», π0 = 14.78π΄ ππΏπ ππ‘ + 0.677πΏπ = π£(π‘) βΉ πΏπ π  = π(π ) π  + 0.677 The voltage across the inductor about the equilibrium point π£ πΏ π‘ = πΏ π(π0 + πΏπ) ππ‘ = πΏ ππΏπ ππ‘ βΉ ππΏ π  = πΏπ πΏπ π  = π πΏπ π  The voltage across the inductor about the equilibrium point ππΏ π  = π  π(π ) π  + 0.677 The final transfer function ππΏ π  π(π ) = π  π  + 0.677 for small excursions about π = 14.78π΄ or, equivalently, about π£ π‘ = 0 System Dynamics and Control 2.162 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
28. 28. 1/11/2016 28 Β§11.Linearization Skill-Assessment Ex.2.13 Problem Find the linearized transfer function, πΊ π  = π(π )/πΌ(π ), for the electrical network. The network contains a nonlinear resistor whose voltage-current relationship is defined by π π = π π£ π. The current source, π(π‘), is a small- signal generator Solution The nodal equation πΆ ππ£ ππ‘ + π π β 2 = π π‘ But πΆ = 1, π£ = π£0 + πΏπ£, π π = π π£ π = π π£ = π π£0+πΏπ£ π(π£0 + πΏπ£) ππ‘ + π π£0+πΏπ£ β 2 = π π‘ System Dynamics and Control 2.163 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien π(π£0+πΏπ£) ππ‘ + π π£0+πΏπ£ β 2 = π π‘ Β§11.Linearization Linearize π π£ π π£ = π π£0 + ππ ππ£ π£=π£0 (π£ β π£0) π π£ = π π£ = π π£0+πΏπ£ π π£0 = π π£0 π£ β π£0 = πΏπ£ ππ ππ£ π£=π£0 = ππ π£ ππ£ π£=π£0 = π π£ π£=π£0 = π π£0 βΉ π π£0+πΏπ£ = π π£0 + π π£0 πΏπ£ The linearized equation ππΏπ£ ππ‘ + π π£0 + π π£0 πΏπ£ β 2 = π π‘ System Dynamics and Control 2.164 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§11.Linearization Setting π π‘ = 0 and letting the circuit reach steady state, the capacitor acts like an open circuit. Thus, π£0 = π£π with π π = 2. But, π π = π π£ π or π£π = lnπ π. Hence, π£0 = ln2 = 0.693 ππΏπ£ ππ‘ + π π£0 + π π£0 πΏπ£ β 2 = π π‘ βΉ ππΏπ£ ππ‘ + 2πΏπ£ = π π‘ Taking the Laplace transform π  + 2 πΏπ£ π  = πΌ(π ) The transfer function πΏπ£(π ) πΌ(π ) = π(π ) πΌ(π ) = 1 π  + 2 about equilibrium System Dynamics and Control 2.165 Modeling in Frequency Domain HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Β§12.Case Studies System Dynamics and Control 2.166 Modeling in Frequency Domain