8/25/2013 System Dynamics 3.01 Solution Methods for Dynamic Models System Dynamics Nguyen Tan Tien 3.03 Solution Methods for Dynamic Models §1.Differential Equations 1.Initial Condition 2𝑥 𝑡 + 6𝑥 𝑡 = ⟹ 𝑥 𝑡 = 𝐶𝑒 −3𝑡 + 0.5 𝐶: constant, can be derived from the value of 𝑥 𝑡0 = 𝑥 𝑡=𝑡0 𝑥(𝑡0 ): initial condition 2.Classification of Differential Equations - Linear differential equation 𝑥 + 3𝑥 = + 𝑡 , 𝑥 + 3𝑡 𝑥 = 5, 3𝑥 + 7𝑥 + 2𝑡 𝑥 = 𝑠𝑖𝑛𝑡 - Nonlinear differential equation 3𝑥 + 6𝑥 = + 𝑡 , 3𝑥 + 5𝑥 + 8𝑥 = 4, 𝑥 + 4𝑥𝑥 + 3𝑥 = - Variable coefficient differential equation 𝑥 + 3𝑡 𝑥 = - Constant coefficient differential equation 𝑥 + 2𝑥 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 3.05 Solution Methods for Dynamic Models §1.Differential Equations - Example 3.1.1 Separation of Variables for a Linear Equation Use separation of variables to solve the following problem for 𝑡 ≥ 0: 𝑥 + 2𝑥 = 20, 𝑥 = Solution 𝑑𝑥 + 2𝑥 𝑡 = 20 ⟹ 𝑑𝑡 𝑥 𝑡 𝑑𝑥 = −2 𝑥 𝑡 − 10 ⟹ ln[𝑥 𝑡 − 10] 𝑥(𝑡) 𝑡 𝑑𝑡 = −2 𝑡 𝑡 ln 𝑥 𝑡 − 10 − ln − 10 𝑥 𝑡 − 10 ⟹ ln = −2𝑡 −7 ⟹ 𝑥 𝑡 = 10 − 7𝑒 −2𝑡 ⟹− HCM City Univ of Technology, Faculty of Mechanical Engineering 3.02 Solution Methods for Dynamic Models §1.Differential Equations - An ordinary differential equation (ODE): an equation containing ordinary, but not partial, derivatives of the dependent variable - The subject of system dynamics is time-dependent behavior, the independent variable in our ODEs will be time 𝑡 3𝑥 + 7𝑥 + 2𝑡 𝑥 = + 𝑠𝑖𝑛𝑡 + 𝑠𝑖𝑛𝑡 : the input or forcing function 𝑥(𝑡): the response - If there is no input (right-hand side is zero), the equation is said to be homogeneous; otherwise, it is nonhomogeneous Solution Methods for Dynamic Models HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics =𝑡−0 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.04 Nguyen Tan Tien Solution Methods for Dynamic Models §1.Differential Equations - The order of differential equation: the order of the highest derivative of the dependent variable in the equation 3𝑥 + 7𝑥 + 2𝑥 = 5: second order differential equation - A model can consist of more than one equation 3𝑥1 + 5𝑥1 − 7𝑥2 = 𝑥2 + 4𝑥1 + 6𝑥2 = 3.Separation of Variables 𝑥=𝑔 𝑡 𝑓 𝑥 𝑑𝑥 ⟹ = 𝑔 𝑡 𝑑𝑡 𝑓 𝑥 𝑥(𝑡) ⟹ 𝑥(0) 𝑑𝑥 = 𝑓(𝑥) 𝑡 𝑔 𝑡 𝑑𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.06 Nguyen Tan Tien Solution Methods for Dynamic Models §1.Differential Equations 4.Trial Solution Method Consider 𝑥 + 𝑎𝑥 = 𝑏, 𝑡 ≥ ⟹ Find the solution in the form 𝑥 𝑡 = 𝐶 + 𝐷𝑒 𝑠𝑡 Subtituiting the solution into the equation 𝑥 + 𝑎𝑥 = 𝑠𝐷𝑒 𝑠𝑡 + 𝑎 𝐶 + 𝐷𝑒 𝑠𝑡 ⟹ 𝑠 + 𝑎 𝐷𝑒 𝑠𝑡 + 𝑎𝐶 = 𝑏 Find 𝑠, 𝐶 𝑠 = −𝑎 𝑠+𝑎 = ⟹ 𝐶 = 𝑏/𝑎 𝑎𝐶 = 𝑏 Find 𝐷 from the initial condition 𝑥 = 𝐶 + 𝐷𝑒 = 𝐶 + 𝐷 ⟹ 𝐷 = 𝑥 − 𝐶 Therefore 𝑏 𝑏 𝑥 𝑡 = + 𝑥 − 𝑒 −𝑎𝑡 𝑎 𝑎 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 3.07 Solution Methods for Dynamic Models §1.Differential Equations - Example 3.1.2 Two Distinct, Real Roots Use trial solution method to solve the following problem for 𝑡 ≥ 0: 𝑥 + 7𝑥 + 10𝑥 = 20, 𝑥 = 5, 𝑥 = Solution Substituiting the trial solution 𝑥 𝑡 = 𝐶 + 𝐷𝑒 𝑠𝑡 into the equation 𝑠 + 7𝑠 + 10 𝐷𝑒 𝑠𝑡 + 10𝐶 = 20 ⟹ 𝑠 + 7𝑠 + 10 = 10𝐶 = 20 ⟹ 𝑠 = −2, 𝑠 = −5, 𝐶 = There are two solutions ⟹ the trial form is not suitable Need an additional term with an arbitrary constant ⟹ The appropriate trial-solution form 𝑥 𝑡 = 𝐶 + 𝐷1 𝑒 𝑠1 𝑡 +𝐷2 𝑒 𝑠2 𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.09 Nguyen Tan Tien Solution Methods for Dynamic Models §1.Differential Equations - Example 3.1.3 Two Repeated, Real Roots Use trial solution method to solve the following problem for 𝑡 ≥ 0: 5𝑥 + 20𝑥 + 20𝑥 = 28, 𝑥 = 5, 𝑥 = Solution Trial solution form 𝑥 𝑡 = 𝐶 + 𝐷𝑒 𝑠𝑡 Substituting this form into the ODE 5𝑠 + 20𝑠 + 20 𝐷𝑒 𝑠𝑡 + 20𝐶 = 28 ⟹ 5𝑠 + 20𝑠 + 20 = 20𝐶 = 28 ⟹ 𝑠 = −2, 𝑠 = −2, 𝐶 = 1.4 Need an additional term with an arbitrary constant ⟹ The appropriate trial-solution form 𝑥 𝑡 = 𝐶 + (𝐷1 + 𝐷2 𝑡)𝑒 𝑠1 𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.11 Nguyen Tan Tien Solution Methods for Dynamic Models §1.Differential Equations - Example 3.1.4 Two Imaginary Roots Use trial solution method to solve the following problem for 𝑡 ≥ 0: 𝑥 + 16𝑥 = 144, 𝑥 = 5, 𝑥 = 12 Solution Trial solution form 𝑥 𝑡 = 𝐶 + 𝐷𝑒 𝑠𝑡 Substituting this form into the ODE 𝑠 + 16 𝐷𝑒 𝑠𝑡 + 16𝐶 = 144 ⟹ 𝑠 + 16 = 16𝐶 = 144 ⟹ 𝑠 = −4𝑗, 𝑠 = +4𝑗, 𝐶 = Therefore 𝑥 𝑡 = 𝐶 + 𝐷1 𝑒 4𝑗𝑡 + 𝐷2 𝑒 −4𝑗𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.08 Solution Methods for Dynamic Models §1.Differential Equations The appropriate trial-solution form 𝑥 𝑡 = 𝐶 + 𝐷1 𝑒 𝑠1 𝑡 +𝐷2 𝑒 𝑠2 𝑡 Substituting this form into the ODE 𝑠12 + 7𝑠1 + 10 𝐷1𝑒𝑠1𝑡 + 𝑠22 + 7𝑠2 + 10 𝐷2𝑒𝑠2𝑡 + 10𝐶 = 20 ⟹ 𝑠12 + 7𝑠1 + 10 = 𝑠1 = −2 𝑠22 + 7𝑠2 + 10 = ⟹ 𝑠2 = −5 𝐶=2 10𝐶 = 20 The solution 𝑥 𝑡 = + 𝐷1𝑒 −2𝑡 +𝐷2 𝑒 −5𝑡 𝐷1, 𝐷2 are calculated from the initial conditions 𝑥 = + 𝐷1 + 𝐷2 = 𝐷 = +6 ⟹ 𝐷2 = −3 𝑥 = −2𝐷1 − 5𝐷2 = ⟹ 𝑥 𝑡 = + 6𝑒 −2𝑡 − 3𝑒 −5𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.10 Nguyen Tan Tien Solution Methods for Dynamic Models §1.Differential Equations The appropriate trial-solution form 𝑥 𝑡 = 𝐶 + (𝐷1 + 𝐷2 𝑡)𝑒 𝑠1 𝑡 Substituting this form into the ODE 5𝑠12 + 20𝑠1 + 20 𝐷1 𝑒 𝑠1 𝑡 + 5𝑠12 + 20𝑠1 + 20 𝐷2 𝑡𝑒 𝑠1 𝑡 + 10𝑠1 + 20 𝐷2𝑒 𝑠1 𝑡 + 20𝐶 = 28 5𝑠12 + 20𝑠1 + 20 = 10𝑠1 + 20 = ⟹ 𝑠1 = −2, 𝐶 = 1.4 20𝐶 = 28 The solution 𝑥 𝑡 = 1.4 + (𝐷1 + 𝐷2 𝑡)𝑒 −2𝑡 𝐷1, 𝐷2 are calculated from the initial conditions 𝐷 = 3.6 𝑥 = 1.4 + 𝐷1 = ⟹ 𝐷2 = 15.2 𝑥 = −2𝐷1 + 𝐷2 = ⟹ ⟹ 𝑥 𝑡 = 1.4 + (3.6 + 15.2𝑡)𝑒 −2𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.12 Nguyen Tan Tien Solution Methods for Dynamic Models §1.Differential Equations The appropriate trial-solution form 𝑥 𝑡 = 𝐶 + 𝐷1 𝑒 4𝑗𝑡 + 𝐷2 𝑒 −4𝑗𝑡 Using Euler’s identities 𝑒𝑗𝜃 = 𝑐𝑜𝑠𝜃 + 𝑗𝑠𝑖𝑛𝜃, 𝜃 = 𝜔𝑡 𝑒 4𝑗𝑡 = 𝑐𝑜𝑠4𝑡 + 𝑗𝑠𝑖𝑛4𝑡 𝑒 −4𝑗𝑡 = 𝑐𝑜𝑠4𝑡 − 𝑗𝑠𝑖𝑛4𝑡 Substituting this form into the solution 𝑥 𝑡 = 𝐶 + 𝐷1 + 𝐷2 𝑐𝑜𝑠4𝑡 + 𝑗 𝐷1 − 𝐷2 𝑠𝑖𝑛4𝑡 or 𝑥 𝑡 = 𝐶 + 𝐵1 𝑐𝑜𝑠4𝑡 + 𝐵2 𝑠𝑖𝑛4𝑡 Evaluating 𝑥(𝑡) and 𝑥(𝑡) at 𝑡 = 𝑥 = 𝐶 + 𝐵1 = 𝐵 = −4 ⟹ 𝐵2 = +3 𝑥 = 4𝐵2 = 12 The solution 𝑥 𝑡 = + 3𝑠𝑖𝑛4𝑡 − 4𝑐𝑜𝑠4𝑡 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 3.13 §1.Differential Equations - Example 3.1.5 Solution Methods for Dynamic Models Motion of a Robot-Arm Link The equation of motion 0.23 + 0.5𝑚𝐿2 𝜃 = 𝑇𝑚 − 4.9𝑚𝐿𝑠𝑖𝑛𝜃 Solve the equation for the case 𝑇𝑚 = 0.5𝑁𝑚,𝑚 = 10𝑘𝑔,𝐿 = 0.3𝑚 Assume the system starts from rest at 𝜃 = and the angle 𝜃 remains small Solution For small 𝜃, 𝑠𝑖𝑛𝜃 ≈ 𝜃 ⟹ 0.23 + 0.5𝑚𝐿2 𝜃 = 𝑇𝑚 − 4.9𝑚𝐿𝜃 0.68𝜃 + 14.7𝜃 = 0.5 0.5 ⟹𝜃 𝑡 = − 𝑐𝑜𝑠4.65𝑡 = 0.034(1 − 𝑐𝑜𝑠4.65𝑡) 14.7 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Solution Methods for Dynamic Models §1.Differential Equations 5.Sumary of the trial solution method Ordinary Differential Equation Solution form 𝑏 𝑥 𝑡 = + 𝐶𝑒 −𝑎𝑡 𝑥 + 𝑎𝑥 = 𝐵, 𝑎 ≠ 𝑥 + 𝑎𝑥 + 𝑏𝑥 = 𝑐, 𝑏 ≠ 𝑎2 > 4𝑏 real roots, distinct 𝑠1 ≠ 𝑠2 𝑎2 = 4𝑏 real roots, repeated 𝑠1 = 𝑠2 𝑎 = 0, 𝑏 > imaginary roots 𝑎 𝑥 𝑡 = 𝐶1 𝑒 𝑠1 𝑡 + 𝐶2 𝑒 𝑠2 𝑡 + 𝑥 𝑡 = (𝐶1 + 𝑡𝐶2 )𝑒 𝑠1 𝑡 + 𝑐 𝑏 𝑐 𝑏 𝑥 𝑡 = 𝐶1 𝑠𝑖𝑛𝜔𝑡 + 𝐶2 𝑐𝑜𝑠𝜔𝑡 + 𝑐 𝑏 𝑠 = ±𝑗𝜔, 𝜔 = 𝑏 𝑐 𝑎 ≠ 0, 𝑎2 < 4𝑏 complex roots 𝑥 𝑡 = 𝑒𝜎𝑡 𝐶1𝑠𝑖𝑛𝜔𝑡 + 𝐶2𝑐𝑜𝑠𝜔𝑡 + 𝑏 𝑠 = 𝜎 ± 𝑗𝜔, 𝜎 = −𝑎/2 𝜔 = 4𝑏 − 𝑎2 /2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 3.17 Solution Methods for Dynamic Models §2.Response Types and Stability - The solution of 𝑥 + 𝑎𝑥 = 𝑏 𝑏 𝑏 𝑥 𝑡 = + 𝑥 − 𝑒 −𝑎𝑡 𝑎 𝑎 or 𝑥 𝑡 = 𝑥 𝑒 −𝑎𝑡 + 𝑏 − 𝑒 −𝑎𝑡 𝑎 Response • Steady-state: the part of the response that remains with time • Transient: the part of the response that disappears with time • Free: the part of the response that depends on the initial conditions • Forced: the part of the response due to the forcing function 𝑥 + 𝑎𝑥 = 𝐵, 𝑎 ≠ 𝑏 𝑥 𝑡 = 𝑎 + 𝐶𝑒 −𝑎𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering 3.14 Solution Methods for Dynamic Models §1.Differential Equations - Example 3.1.6 Two Complex Roots Use trial solution method to solve the following problem for 𝑡 ≥ 0: 𝑥 + 6𝑥 + 34𝑥 = 68, 𝑥 = 5, 𝑥 = Solution Subtituting 𝑥 𝑡 = 𝐶 + 𝐷𝑒 𝑠𝑡 into the ODE 𝑠 + 6𝑠 + 34 𝐷𝑒 𝑠𝑡 + 34𝐶 = 68 𝑠 + 6𝑠 + 34 = ⟹ 𝑠 = −3 ± 5𝑗, 𝐶 = 34𝐶 = 68 Therefore 𝑥 𝑡 = 𝐶 + 𝐷1 𝑒 (−3+5𝑗)𝑡 + 𝐷2 𝑒 (−3−5𝑗)𝑡 ⟹ 𝑥 𝑡 = 𝐶 + 𝑒 −3𝑡 (𝐷1 𝑒 5𝑗𝑡 + 𝐷2 𝑒 −5𝑗𝑡 ) ⟹ 𝑥 𝑡 = 𝐶 + 𝑒 −3𝑡 (𝐵1 𝑐𝑜𝑠5𝑡 + 𝐵2 𝑠𝑖𝑛5𝑡) ⟹ With the initial conditions 𝑥 𝑡 = + 𝑒 −3𝑡 3𝑐𝑜𝑠5𝑡 + Nguyen Tan Tien 3.15 System Dynamics HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 𝑠𝑖𝑛5𝑡 Nguyen Tan Tien 3.16 Solution Methods for Dynamic Models §1.Differential Equations 6.Assessment of Solution Behavior - The characteristic equation can be quickly identified from ODE by replacing 𝑥 with 𝑠, 𝑥 with 𝑠 , and so forth - Example 3𝑥 + 30𝑥 + 222𝑥 = 148 The characteristic equation 𝑠 + 10𝑠 + 74 = ⟹ 𝑠 − −5 − 7𝑗 𝑠 − −5 + 7𝑗 = The solution in the form 𝑥 𝑡 = 𝑒 −5𝑡 𝐶1 𝑠𝑖𝑛7𝑡 + 𝐶2 𝑐𝑜𝑠7𝑡 + 𝑐 𝑎 ≠ 0, 𝑎2 < 4𝑏 : 𝑠 = 𝜎 ± 𝑗𝜔, 𝜎 = −𝑎/2, 𝜔 = 4𝑏 − 𝑎2/2 𝑥 𝑡 = 𝑒𝜎𝑡 𝐶1 𝑠𝑖𝑛𝜔𝑡 + 𝐶2 𝑐𝑜𝑠𝜔𝑡 + 𝑏 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 3.18 Solution Methods for Dynamic Models §2.Response Types and Stability 1.The time constant The 1st order model 𝑥 + 𝑎𝑥 = 𝑏 ⟹ 𝑥 + 𝑥 = 𝑏, 𝜏 ≡ 1/𝑎 𝜏 give the free response 𝑥 𝑡 = 𝑥 𝑒 −𝑎𝑡 𝑡 ⟹ 𝑥 𝑡 = 𝑥(0)𝑒 − 𝜏 𝜏: time constant • measure of the exponential decay curve • estimate how long it will take for the transient response to disappear 𝑥 + 𝑎𝑥 = 𝑏, 𝑎 ≠ Nguyen Tan Tien 16 𝑏 𝑥 𝑡 = 𝑎 + 𝐶𝑒 −𝑎𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 3.19 Solution Methods for Dynamic Models §2.Response Types and Stability System Dynamics 3.20 Solution Methods for Dynamic Models §2.Response Types and Stability The time constant is useful also for analyzing the response when the forcing function is a constant The total response in terms of 𝜏 by substituting 𝑎 = 1/𝜏 𝑥𝑡 = 𝑏𝜏 steady state 𝑡 𝑡 + 𝑥 − 𝑏𝜏 𝑒− 𝜏 = 𝑥𝑠𝑠 + [𝑥 − 𝑥𝑠𝑠]𝑒− 𝜏 transient state The response approaches the constant value 𝑏𝜏 as 𝑡 → ∞ 𝑥𝑠𝑠 = lim 𝑥(𝑡) = 𝑏𝜏 𝑡→∞ From figure • after 𝑡 = 𝜏, 𝑥(𝑡) has decayed to 37% of its initial value • after 𝑡 = 4𝜏, 𝑥(𝑡) has decayed to 02% of its initial value HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.21 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability System Dynamics 3.23 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability 2.The Dominant-Root Approximation - The time constant concept is not limited to first-order models It can also be used to estimate the response time of higherorder models - Example 3.2.2 Responses for Second-Order, Complex Roots Identify the responses of the equation 𝑥 + 6𝑥 + 34𝑥 = 𝑐 (𝑠 = −3 ± 5𝑗) Solution Following the procedure used in Example 3.1.6 𝑐 𝑥 𝑡 = + 𝑒 −3𝑡 𝐵1 𝑐𝑜𝑠5𝑡 + 𝐵2 𝑠𝑖𝑛5𝑡 34 𝑥 + 𝑎𝑥 + 𝑏𝑥 = 𝑐,𝑎2 < 4𝑏:𝑠 = 𝜎 ± 𝑗𝜔,𝜎 = −𝑎/2,𝜔 = 4𝑏 − 𝑎2/2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.22 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability - Example 3.2.1 Responses for Second-Order, Distinct Roots Identify the responses of 𝑥 + 7𝑥 + 10𝑥 = 𝑐 (roots: −2, −5) Solution 𝑐 The solution 𝑥 𝑡 = + 𝐷1𝑒 −2𝑡 + 𝐷2 𝑒 −5𝑡 10 Submit to ODE 𝑐 𝑐 ⟹ 𝐷1 = 𝑥 + 𝑥 − , 𝐷2 = − 𝑥 − 𝑥 + 3 3 15 Arranging the solution in another form 𝑥 𝑡 = 𝑥 + 𝑥 𝑒 −2𝑡 + − 𝑥 − 𝑥 𝑒 −5𝑡 3 3 1 −2𝑡 +𝑐 − 𝑒 10 The forced response (for which 𝑥(0) = 0) is plotted in the figure • at 𝑡 = 𝜏, the response is 63% of the steady-state value • at 𝑡 = 4𝜏, the response is 98% of the steady-state value • at 𝑡 = 5𝜏, the response is 99% of the steady-state value For most engineering purposes: 𝑥(𝑡) reaches steady-state at 𝑡 = 4𝜏 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.24 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability The coefficients 𝐵1 and 𝐵2 can be found in terms of arbitrary initial conditions in the usual way 𝑐 34𝑥 + 102𝑥 − 3𝑐 𝐵1 = 𝑥 − , 𝐵2 = 34 170 𝑥 + 3𝑥(0) −3𝑡 ⟹ 𝑥 𝑡 = 𝑥 𝑒 −3𝑡 𝑐𝑜𝑠5𝑡 + 𝑒 𝑠𝑖𝑛5𝑡 + 𝑐 − 𝑒 −3𝑡 𝑐𝑜𝑠5𝑡 − 𝑒 −3𝑡 𝑠𝑖𝑛5𝑡 34 Model’s time constant: τ = 1/3 ⟹ the response is essentially at steady state for 𝑡 > 4τ = 4/3 𝑐 𝑥 𝑡 = 𝑒𝜎𝑡 𝐶1𝑠𝑖𝑛𝜔𝑡 + 𝐶2𝑐𝑜𝑠𝜔𝑡 + 𝑏 Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 3.25 Solution Methods for Dynamic Models §2.Response Types and Stability 3.Time Constants and Complex Roots - Example 3.2.3 Responses for Second-Order, Imaginary Roots Identify the responses of 𝑥 + 16𝑥 = 𝑐 Solution The characteristic roots 𝑠 = ±4𝑗, the solution form 𝑐 𝑥 𝑡 = + 𝐵1 𝑐𝑜𝑠4𝑡 + 𝐵2 𝑠𝑖𝑛4𝑡 16 no terms that disappear as 𝑡 → ∞ ⟹ no transient response The free and forced responses 𝑥(0) 𝑐 𝑥 𝑡 = 𝑥 𝑐𝑜𝑠4𝑡 + 𝑠𝑖𝑛4𝑡 + (1 − 𝑐𝑜𝑠4𝑡) 16 𝑥 + 𝑎 𝑥 + 𝑏𝑥 = 𝑐: 𝑠 = ±𝑗𝜔, 𝜔 = 𝑏 𝑥 𝑡 = 𝐶1 𝑠𝑖𝑛𝜔𝑡 + 𝐶2 𝑐𝑜𝑠𝜔𝑡 + HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.27 𝑐 𝑏 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability - The stability properties of a linear model are determined from its characteristics roots - The first-order model 𝑥 + 𝑎𝑥 = 𝑓(𝑡) The characteristic equation 𝑠 + 𝑎 = The free response 𝑥 𝑡 = 𝑥(0)𝑒 −𝑎𝑡 𝑠 = −𝑎 < 0: 𝑡 ⟶ ∞, 𝑥(𝑡) ⟶ 0: the model is stable 𝑠 = −𝑎 > 0: 𝑡 ⟶ ∞, 𝑥(𝑡) ⟶ ∞ : the model is unstable 𝑠 = 𝑎 = 0: 𝑡 ⟶ ∞, 𝑥(𝑡) ↛ 0, ∞: the model is neural stability System Dynamics 3.26 Solution Methods for Dynamic Models §2.Response Types and Stability 4.Stability - Unstable: the free response approaches ∞ as 𝑡 → ∞ - Stable: the free response approaches - Neutral stability: the borderline between stable and unstable The free response does not approach both ∞ and HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.28 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability - The second-order models with the same initial conditions 𝑥 = 1; 𝑥 = 1 𝑥 − 4𝑥 = 𝑓(𝑡) 𝑠 = ±2 + 0𝑗 𝑥 𝑡 = 𝑒 2𝑡 + 𝑒 −2𝑡 2 𝑥 − 4𝑥 + 229𝑥 = 𝑓(𝑡) 𝑠 = +2 ± 15𝑗 𝑥 𝑡 = 𝑒2𝑡 𝑐𝑜𝑠15𝑡 − 𝑠𝑖𝑛15𝑡 15 𝑥 + 256𝑥 = 𝑓(𝑡) 𝑠 = ± 16𝑗 𝑥 𝑡 = 𝑐𝑜𝑠16𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.29 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability - Stability Test for Linear Constant-Coefficient Models • A constant-coefficient linear model is stable if and only if all of its characteristic roots have negative real parts • The model is neutrally stable if one or more roots have a zero real part, and the remaining roots have negative real parts • The model is unstable if any root has a positive real part HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.30 Nguyen Tan Tien Solution Methods for Dynamic Models §2.Response Types and Stability 5.A Physical Example - Pendulum motion equation (𝜃 ≈ 0) 𝑚𝐿2 𝜃 + 𝑐𝜃 + 𝑚𝑔𝐿𝜃 = - Equilibrium point: 𝜃 = −𝑐 ± 𝑐 − 4𝑚2 𝐿3 𝑔 2𝑚𝐿2 • 2𝑚𝐿 𝐿𝑔 > 𝑐 > 0: damped oscillating Roots 𝑠= • 𝑐 > 2𝑚𝐿 𝐿𝑔: no oscillating • 𝑐 = 0: oscillating about 𝜃 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 3.31 Solution Methods for Dynamic Models §2.Response Types and Stability - Pendulum motion equation (𝜃 ≈ 0) 𝑚𝐿2 𝜃 + 𝑐𝜃 + 𝑚𝑔𝐿𝜃 = - Equilibrium point: 𝜃 = 𝜋 −𝑐 ± 𝑐 + 4𝑚2 𝐿3 𝑔 2𝑚𝐿2 - Pendulum will not oscillate about 𝜃 = 𝜋 but will continue to fall away if disturbed ⟹ the system is unstable - The model is based on the assumption that 𝜃 ≈ ⟹ cannot draw any conclusions from the model regarding the behavior when 𝜃 is not near 𝜋 𝑠= Roots HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.33 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.32 Solution Methods for Dynamic Models §2.Response Types and Stability 6.Routh-Hurwitz Condition - The characteristic equation of many systems has the form 𝑚𝑠 + 𝑐𝑠 + 𝑘 = - Routh-Hurwitz condition: the second-order system whose characteristic polynomial is 𝑚𝑠 + 𝑐𝑠 + 𝑘 = is stable if and only if 𝑚, 𝑐, and 𝑘 have the same sign HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.34 Nguyen Tan Tien Fluid and Thermal Systems §2.Response Types and Stability 7.Stability and Equilibrium - An equilibrium: a state of no change The pendulum in the figure is in equilibrium at 𝜃 = and when perfectly balanced at 𝜃 = 𝜋 • the equilibrium at 𝜃 = 0: stable • the equilibrium at 𝜃 = 𝜋: unstable ⟹ the same physical system can have different stability characteristics at different equilibria - Stability is not a property of the system alone, but is a property of a specific equilibrium of the system - When we speak of the stability properties of a model, we are actually speaking of the stability properties of the specific equilibrium on which the model is based §2.Response Types and Stability - The figure shows a ball on a surface that has a valley and a hill HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.35 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method - The Laplace transforms the problem in time-domain to problem in s-domain, then applying the solution in s-domain, and finally using inverse transform to converse the solution back to the time-domain - The bottom of the valley is an equilibrium, and if the ball is displaced slightly from this position, it will • oscillate forever about the bottom if there is no friction: neutrally stable • return to the bottom if friction is present: stable - If displace the ball so much to the left that it lies outside the valley, it will never return: locally stable but globally unstable - If the system returns to its equilibrium for any initial displacement: globally stable - The equilibrium on the hilltop is globally unstable System Dynamics 3.36 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method - Notation for the Laplace and inverse Laplace transforms 𝑋 𝑠 =ℒ 𝑥 𝑡 𝑥 𝑡 = ℒ −1 {𝑋(𝑠)} - Table of Laplace transform pairs - The Laplace transform ℒ{𝑥 𝑡 } of a function 𝑥(𝑡) is defined as 𝑇 ℒ 𝑥 𝑡 𝑥 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = lim 𝑇→∞ but is usually expressed more compactly as ∞ ℒ 𝑥 𝑡 𝑥 𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 8/25/2013 System Dynamics 3.37 Fluid and Thermal Systems §3.The Laplace Transform Method System Dynamics 3.38 Fluid and Thermal Systems §3.The Laplace Transform Method 1.Transforms of Common Functions - Example 3.3.1 Transform of a Constant Suppose 𝑥(𝑡) = 𝑐 , a constant, for 𝑡 ≥ Determine its Laplace transform Solution From the transform definition 𝑇 𝑇 𝑐𝑒 −𝑠𝑡 𝑑𝑡 = 𝑐 lim ℒ 𝑥(𝑡) = lim 𝑇→∞ −𝑠𝑡 𝑒 −𝑠 = 𝑐 lim 𝑇→∞ 𝑒 −𝑠𝑡 𝑑𝑡 𝑇→∞ 𝑇 −𝑠𝑇 −𝑠×0 𝑐 ⟹ ℒ 𝑐 = 𝑐 lim 𝑒 − 𝑒 = 𝑇→∞ −𝑠 −𝑠 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.39 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method - The step function HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.40 𝑇 ℒ 𝑒 −𝑎𝑡 = lim 𝑇→∞ 𝑡0 HCM City Univ of Technology, Faculty of Mechanical Engineering = lim 𝑒 − 𝑠+𝑎 −(𝑠 + 𝑎) 𝑇→∞ 341 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method 2.Properties of the Laplace Transform 𝑒 −(𝑠+𝑎)𝑡 𝑑𝑡 𝑡 𝑇 − 𝑒 − 𝑠+𝑎 ×0 𝑠+𝑎 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 𝑇→∞ 𝑇 𝑒 − 𝑠+𝑎 − 𝑠+𝑎 𝑇→∞ = 𝑀/𝑠 𝑇 𝑒 −𝑎𝑡 𝑒 −𝑠𝑡 𝑑𝑡 = lim = lim ⟹ ℒ 𝑒 −𝑎𝑡 = System Dynamics Fluid and Thermal Systems §3.The Laplace Transform Method - Example 3.3.2 The Exponential Function Derive the Laplace transform of the exponential function 𝑥(𝑡) = 𝑒 −𝑎𝑡 , where 𝑎 is a constant Solution From the transform definition Unit-step function Step function 𝑥(𝑡) = 𝑀𝑢𝑠 (𝑡) 𝑋 𝑠 = ℒ 𝑀𝑢𝑠 𝑡 Nguyen Tan Tien 3.42 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method - Example 3.3.3 The Sine and Cosine Functions Derive the Laplace transforms of 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 and 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡, where 𝑎 and 𝜔 are constants Solution Recall: - the Euler identity 𝑒𝑗𝜃 = 𝑐𝑜𝑠𝜃 + 𝑗𝑠𝑖𝑛𝜃, with 𝜃 = 𝜔𝑡 - the relation 𝑥 + 𝑗𝑦 𝑥 + 𝑗𝑦 = = 𝑥 − 𝑗𝑦 (𝑥 − 𝑗𝑦)(𝑥 + 𝑗𝑦) 𝑥 + 𝑦 We have 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 + 𝑗𝑠𝑖𝑛𝜔𝑡 = 𝑒 −𝑎𝑡 𝑒𝑗𝜔𝑡 = 𝑒 − 𝑎−𝑗𝜔 𝑡 𝑠 + 𝑎 + 𝑗𝜔 ℒ 𝑒 −(𝑎−𝑗𝜔)𝑡 = = 𝑠 + (𝑎 − 𝑗𝜔) (𝑠 + 𝑎)2 + (𝜔)2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien HCM City Univ of Technology, Faculty of Mechanical Engineering (1) Nguyen Tan Tien 8/25/2013 System Dynamics 3.43 Fluid and Thermal Systems §3.The Laplace Transform Method 𝑠+𝑎 𝜔 ℒ 𝑒 −(𝑎−𝑗𝜔)𝑡 = +𝑗 (𝑠 + 𝑎)2 +𝜔 (𝑠 + 𝑎)2 +𝜔 from eq.(1) ℒ 𝑒 −(𝑎−𝑗𝜔)𝑡 = ℒ 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 + 𝑗𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 System Dynamics ∞ Fluid and Thermal Systems System Dynamics 3.47 ∞ 𝑡𝑥 𝑡 𝑒−𝑠𝑡𝑑𝑡 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method b From the time-shifting property 𝑥(𝑡)𝑒 −(𝑠+𝑎)𝑡 𝑑𝑡 = −𝑎𝑡 ⟹ ℒ 𝑒 𝑥(𝑡) = 𝑋(𝑠 + 𝑎) where 𝑋 𝑠 = ℒ 𝑥(𝑡) - Example 3.3.4 The Function 𝑡𝑒 −𝑎𝑡 Derive the Laplace transform of the function 𝑡𝑒 −𝑎𝑡 Solution 𝑥 𝑡 =𝑡⟹𝑋 𝑠 = 𝑠 1 ℒ 𝑡𝑒 −𝑎𝑡 = ℒ 𝑒 −𝑎𝑡 𝑥(𝑡) = = 𝑠 𝑠→𝑠+𝑎 (𝑠 + 𝑎)2 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 3.46 Fluid and Thermal Systems 𝑢𝑠 (𝑡 − 𝐷) is called the shifted step function Determine 𝑋(𝑠) Solution 𝑇 𝑇→∞ 𝐷 𝑀𝑢𝑠(𝑡 − 𝐷)𝑒−𝑠𝑡𝑑𝑡 = lim ℒ 𝑥(𝑡) = lim 𝑇→∞ 𝑇 × 𝑒−𝑠𝑡𝑑𝑡 + 𝑀𝑒−𝑠𝑡𝑑𝑡 𝐷 𝑇 −𝑠𝑡 −𝑠𝑇 −𝑠𝐷 𝑒 = 𝑀 lim 𝑒 − 𝑒 𝑇→∞ −𝑠 −𝑠 −𝑠 𝐷 𝑀 ⟹ ℒ 𝑥(𝑡) = 𝑒 −𝑠𝐷 𝑠 and ℒ 𝑢𝑠 (𝑡 −Faculty 𝐷) = 𝑒 −𝑠𝐷 /𝑠 HCM City Univ of Technology, of Mechanical Engineering Nguyen Tan Tien = + 𝑀 lim 𝑇→∞ System Dynamics 3.48 Fluid and Thermal Systems §3.The Laplace Transform Method 3.The Derivative Property Applying integration by parts to the definition of the transform ∞ 𝑑𝑥 𝑑𝑥 −𝑠𝑡 ℒ = 𝑒 𝑑𝑡 𝑑𝑡 𝑑𝑡 = 𝑥 𝑡 𝑒 −𝑠𝑡 The pulse = A unit-step + A shifted, negative unit-step 𝑃(𝑡) = 𝑢𝑠 (𝑡) − 𝑢𝑠 (𝑡 − 𝐷) From the time-shifting property 𝑃 𝑠 = ℒ 𝑢𝑠 𝑡 − ℒ 𝑢𝑠 𝑡 − 𝐷 1 = − 𝑒−𝑠𝐷 𝑠 𝑠 ⟹ ℒ 𝑃 𝑡 = − 𝑒 −𝑠𝐷 𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑒 −𝑎𝑡 𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 §3.The Laplace Transform Method - Example 3.3.6 The Shifted Step Function If the discontinuity in the unit-step function occurs at 𝑡 = 𝐷 𝑡𝐷 - Example 3.3.5 The Function 𝑡𝑐𝑜𝑠𝜔𝑡 Derive the Laplace transform of the function 𝑡𝑐𝑜𝑠𝜔𝑡 Solution 𝑠 𝑥 𝑡 = 𝑐𝑜𝑠𝜔𝑡 ⟹ 𝑋 𝑠 = 𝑠 + 𝜔2 𝑑 𝑠 𝑠 − 𝜔2 ℒ 𝑡𝑐𝑜𝑠𝜔𝑡 = ℒ 𝑡𝑥(𝑡) = − = 𝑑𝑠 𝑠 + 𝜔 𝑠 + 𝜔2 HCM City Univ of Technology, Faculty of Mechanical Engineering Fluid and Thermal Systems ∞ ℒ 𝑒 −𝑎𝑡 𝑥(𝑡) = Nguyen Tan Tien 3.45 §3.The Laplace Transform Method - Multiplication by 𝑡 ∞ 𝑑 𝑑 𝑋𝑠 = 𝑥 𝑡 𝑒−𝑠𝑡 𝑑𝑡 = − 𝑑𝑠 𝑑𝑠 = −ℒ{𝑡𝑥(𝑡)} 𝑑𝑋(𝑠) ⟹ ℒ 𝑡𝑥(𝑡) = − 𝑑𝑠 3.44 §3.The Laplace Transform Method - Shifting along the 𝑠-axis or multiplication by an exponential Comparing the above two equation, the Laplace transforms of the exponentially decaying sine and cosine functions are 𝑠+𝑎 ℒ 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 = (𝑠 + 𝑎)2 +𝜔 𝜔 ℒ 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 = (𝑠 + 𝑎)2 +𝜔 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien ∞ ∞ 𝑥(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 +𝑠 = 𝑠ℒ 𝑥 𝑡 − 𝑥 = 𝑠𝑋 𝑠 − 𝑥(0) Extend to higher derivatives 𝑑2𝑥 ℒ = 𝑠 𝑋 𝑠 − 𝑠𝑥 − 𝑥(0) 𝑑𝑡 ℒ 𝑑𝑛 𝑥 = 𝑠𝑛𝑋 𝑠 − 𝑑𝑡 𝑛 𝑛 𝑠 𝑛−𝑘 𝑔𝑘−1 , 𝑘=1 HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑔𝑘−1 = 𝑑 𝑘−1 𝑥 𝑑𝑡𝑘−1 𝑡=0 Nguyen Tan Tien 8/25/2013 System Dynamics 3.49 Fluid and Thermal Systems §3.The Laplace Transform Method 4.The Initial Value Theorem Use to find the value of the function 𝑥(𝑡) at 𝑡 = 0+ (a time infinitesimally greater than 0) with given the transform 𝑋(𝑠) 𝑥 0+ = lim 𝑥(𝑡) = lim [𝑠𝑋(𝑠)] 𝑡→0+ 𝑠→∞ System Dynamics HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.51 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method 6.Solving Equations with the Laplace Transform Consider the linear first-order equation 𝑥 + 𝑎𝑥 = 𝑓(𝑡) 𝑓(𝑡): the input 𝑎: a constant Multiply both sides of the equation by 𝑒 −𝑠𝑡 and then integrate ∞ ∞ (𝑥 + 𝑎𝑥)𝑒 −𝑠𝑡 𝑑𝑡 = 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 ℒ 𝑥 + 𝑎𝑥 = ℒ{𝑓(𝑡)} ⟹ ℒ 𝑥 + 𝑎ℒ 𝑥 = ℒ{𝑓(𝑡)} 𝑠𝑋 𝑠 − 𝑥 + 𝑎𝑋 𝑠 = 𝐹 𝑠 𝑥(0) ⟹𝑋 𝑠 = + 𝐹(𝑠) 𝑠+𝑎 𝑠+𝑎 𝑥(0) ⟹ 𝑥 𝑡 = ℒ −1 + ℒ −1 𝐹(𝑠) 𝑠+𝑎 𝑠+𝑎 or Then free response 𝑡→∞ 𝑠→0 𝑋 𝑠 = (𝑠 + 4)2 +49 7𝑠 =0 + 4)2 +49 This is confirmed by evaluating the inverse transform 𝑥 𝑡 = ℒ −1 𝑋 𝑠 = 𝑒 −4𝑡 sin7𝑡 ⟹ 𝑥 ∞ = 𝑒 −(4×0) sin(7 × 0) = The final value theorem does not apply to a periodic function ⟹ 𝑥 ∞ = lim 𝑠𝑋(𝑠) = lim 𝑠→0 𝑠→0 (𝑠 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.52 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method - Example 3.3.8 Step Response of a First-Order Equation Suppose that the input 𝑓(𝑡) of the equation 𝑥 + 𝑎𝑥 = 𝑓(𝑡) is a step function of magnitude 𝑀 whose transform is 𝐹(𝑠) = 𝑀/𝑠 Obtain the expression for the complete response Solution The forced response is obtained from 1 𝑀 𝑀 1 𝑥 𝑡 = ℒ−1 𝐹(𝑠) = ℒ−1 = ℒ−1 − 𝑠 +𝑎 𝑠+𝑎 𝑠 𝑎 𝑠 𝑠+𝑎 𝑀 ⟹𝑥 𝑡 = − 𝑒 −𝑎𝑡 𝑎 The complete response 𝑀 𝑥 𝑡 = 𝑥 𝑒 −𝑎𝑡 + − 𝑒 −𝑎𝑡 𝑎 forced response HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Fluid and Thermal Systems Example Example 7𝑠 + 𝑋 𝑠 = 𝑠(𝑠 + 6) 7𝑠 + ⟹ 𝑥 0+ = lim 𝑠 =7 𝑠→∞ 𝑠(𝑠 + 6) This is confirmed by evaluating the inverse transform 20 𝑥 𝑡 = ℒ −1 𝑋 𝑠 = + 𝑒 −6𝑡 3 20 20 ⟹ 𝑥 = + 𝑒 −6×0 = + =7 3 3 3.50 §3.The Laplace Transform Method 5.The Final Value Theorem Use to find the limit of the function 𝑥(𝑡) as 𝑡 → ∞ 𝑓 ∞ = lim 𝑓(𝑡) = lim 𝑠𝐹(𝑠) 3.53 Nguyen Tan Tien Fluid and Thermal Systems HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.54 §3.The Laplace Transform Method - Example 3.3.9 Ramp Response of a First-Order Equation Determine the complete response of the following model, which has a ramp input 𝑥 + 3𝑥 = 5𝑡, 𝑥 = 10 Solution Applying the transform to the equation we obtain 𝑠𝑋 𝑠 − 𝑥 + 3𝑋 𝑠 = 𝑠 𝑥(0) 10 ⟹𝑋 𝑠 = + = + 𝑠 + 𝑠 (𝑠 + 3) 𝑠 + 𝑠 (𝑠 + 3) Express the second term on the right as 𝐶1 𝐶2 𝐶3 𝐶1 𝑠 + + 𝐶2𝑠 𝑠 + + 𝐶3𝑠2 ≡ + + = 𝑠2(𝑠 + 3) 𝑠2 𝑠 𝑠 + 𝑠2(𝑠 + 3) 𝐶2 + 𝐶3 𝑠 + 𝐶1 + 3𝐶2 𝑠 + 3𝐶1 = 𝑠 (𝑠 + 3) §3.The Laplace Transform Method Comparing the numerators 𝐶2 + 𝐶3 = 𝐶1 = 5/3 𝐶1 + 3𝐶2 = ⟹ 𝐶2 = −5/9 3𝐶1 = 𝐶3 = 5/9 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Fluid and Thermal Systems The forced response 5 𝐶1 𝑡 + 𝐶2 + 𝐶3 𝑒 −3𝑡 = 𝑡 − + 𝑒 −3𝑡 9 The complete response 5 𝑥 𝑡 = 10𝑒−3𝑡 + 𝑡 − + 𝑒−3𝑡 9 For 𝑡 > 4/3 , 𝑒 −3𝑡 < 0.02, → the response is approximately given by 𝑥(𝑡) = 5𝑡/3 − 5/9 Nguyen Tan Tien 8/25/2013 System Dynamics 3.55 Fluid and Thermal Systems System Dynamics 3.56 Fluid and Thermal Systems §3.The Laplace Transform Method - Example 3.3.10 Transform Inversion for Complex Factors Invert the following transform 8𝑠 + 13 𝑋 𝑠 = 𝑠 + 4𝑠 + 53 Solution Express 𝑋(𝑠) as a sum of terms 8𝑠 + 13 𝑋𝑠 = (𝑠 + 2)2+72 (𝑠 + 2) =8 − (𝑠 + 2)2+72 (𝑠 + 2)2+72 ⟹ 𝑥 𝑡 = 8𝑒 −2𝑡 𝑐𝑜𝑠7𝑡 − 𝑒 −2𝑡 𝑠𝑖𝑛7𝑡 §3.The Laplace Transform Method - Example 3.3.11 Step Response of a Second-Order Equation Obtain the complete response of the following model HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.57 Nguyen Tan Tien Fluid and Thermal Systems 𝑥 + 4𝑥 + 53𝑥 = 15𝑢𝑠 𝑡 𝑥 = 8, 𝑥 = −19 Solution Transforming the equation gives 𝑠2𝑋 𝑠 − 𝑠𝑥 − 𝑥 + 𝑠𝑋 𝑠 − 𝑥 + 53𝑋 𝑠 = 15 𝑠 Solve for 𝑋(𝑠) using the given initial conditions 𝑥 𝑠 + 𝑥 + 4𝑥(0) 15 𝑋 𝑠 = + 𝑠 + 4𝑠 + 53 𝑠(𝑠 + 4𝑠 + 53) 8𝑠 + 13 15 (1) = + 𝑠 + 4𝑠 + 53 𝑠(𝑠 + 4𝑠 + 53) The first term on the right of eq.(1) corresponds to the free response 8𝑒 −2𝑡 𝑐𝑜𝑠7𝑡 − (3/7)𝑒 −2𝑡 𝑠𝑖𝑛7𝑡 (Ex 3.3.10) (2) System Dynamics 3.58 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method The second term on the right of eq.(1) corresponds to the forced response It can be expressed as follows 15 𝐶1 𝑠+2 = + 𝐶2 + 𝐶3 𝑠[ 𝑠 + 2 + 72 ] 𝑠 𝑠 + 2 + 72 𝑠 + 2 + 72 𝐶1 𝑠 + 2 + 72 + 𝐶2 𝑠 𝑠 + + 7𝐶3𝑠 = 𝑠[ 𝑠 + 2 + 72 ] (𝐶1 + 𝐶2) 𝑠 + 2 + (4𝐶1 +2𝐶2 + 7𝐶3)𝑠 +53𝐶1 = 𝑠[ 𝑠 + 2 + 72 ] Comparing numerators on the left and right sides 𝐶1 + 𝐶2 = 𝐶1 = 15/53 4𝐶1 + 2𝐶2 + 7𝐶3 = ⟹ 𝐶2 = −15/53 53𝐶1 = 15 𝐶3 = −30/371 15 15 −2𝑡 30 −2𝑡 − 𝑒 𝑐𝑜𝑠7𝑡 − 𝑒 𝑠𝑖𝑛7𝑡 (3) The forced response 53 53 371 §3.The Laplace Transform Method The complete response is the sum of the free and forced responses given by equations 15 409 −2𝑡 27 −2𝑡 𝑥 𝑡 = − 𝑒 𝑐𝑜𝑠7𝑡 − 𝑒 𝑠𝑖𝑛7𝑡 (3) 53 53 371 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.59 Nguyen Tan Tien Fluid and Thermal Systems The oscillations are difficult to see for 𝑡 > because 𝑒−2𝑡 < 0.02 for 𝑡 > So for most practical purposes we may say that the response is essentially constant with a value 15/53 for 𝑡 > System Dynamics 3.60 Nguyen Tan Tien Fluid and Thermal Systems §3.The Laplace Transform Method Alternatively, combine the terms on the right side of eq.(1) 𝑥 𝑠 + 𝑥 + 4𝑥 𝑠 + 15 8𝑠 + 13𝑠 + 15 𝑋 𝑠 = = 𝑠(𝑠 + 4𝑠 + 53) 𝑠[ 𝑠 + 2 + 72 ] 𝐶1 𝑠+2 = + 𝐶2 + 𝐶3 𝑠 𝑠 + 2 + 72 𝑠 + 2 + 72 𝐶1 + 𝐶2 𝑠 + 4𝐶1 + 2𝐶2 + 7𝐶3 𝑠 + 53𝐶1 = 𝑠[ 𝑠 + 2 + 72 ] Comparing numerators on the left and right sides 𝐶1 + 𝐶2 = 𝐶1 = 15/53 4𝐶1 + 2𝐶2 + 7𝐶3 = 13 ⟹ 𝐶2 = 409/53 53𝐶1 = 15 𝐶3 = −27/53 The complete response 15 409 −2𝑡 27 −2𝑡 𝑥 𝑡 = − 𝑒 𝑐𝑜𝑠7𝑡 − 𝑒 𝑠𝑖𝑛7𝑡 53 53 371 §3.The Laplace Transform Method Step response of a second-order equation with complex roots Using a transform of the following form 𝐴𝑠2 + 𝐵𝑠 + 𝐶 𝑠+𝑎 𝑏 = 𝐶1 + 𝐶2 + 𝐶3 𝑠[ 𝑠 + 𝑎 + 𝑏2] 𝑠 𝑠 + 𝑎 + 𝑏2 𝑠 + 𝑎 + 𝑏2 where the coefficients 𝐶 𝐵 − 𝑎𝐴 − 𝑎𝐶1 𝐶1 = , 𝐶2 = 𝐴 − 𝐶1 , 𝐶3 = 𝑎 + 𝑏2 𝑏 The response is 𝑥 𝑡 = 𝐶1 + 𝐶2 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝑏𝑡 + 𝐶3 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝑏𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 10 8/25/2013 System Dynamics 3.61 Fluid and Thermal Systems System Dynamics 3.62 Fluid and Thermal Systems §4.Transfer Function - Response of a linear system = Free response + Forced response • To focus the analysis on the effects of the input only by taking the initial conditions to be zero temporarily • Then, the free response due to any nonzero initial conditions can be added to the result - Consider the model 𝑥 + 𝑎𝑥 = 𝑓(𝑡) Transforming both sides of the equation 𝑠𝑋 𝑠 + 𝑎𝑋 𝑠 = 𝐹 𝑠 obtain the transfer function 𝑋(𝑠) 𝑇 𝑠 = = 𝐹(𝑠) 𝑠 + 𝑎 - A Transfer Function is the ratio of the output of a system to the input of a system, in the Laplace domain considering its initial conditions and equilibrium point to be zero §4.Transfer Function The TF concept is extremely useful for several reasons - Transfer Functions and Software • Matlab accepts a description based on the TF • Simulink uses TF as the basis graphical system description called the block diagram - ODE Equivalence The transfer function ⟺ The ODE - The Transfer Function and Characteristic Roots • The characteristic polynomial is the denominator of the TF • The roots of characteristics equation give the information about the stability of the system HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.63 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.64 Nguyen Tan Tien Fluid and Thermal Systems §4.Transfer Function - Example 3.4.1 Two Inputs and One Output Obtain the transfer functions 𝑋(𝑠)/𝐹(𝑠) and 𝑋(𝑠)/𝐺(𝑠) for the following equation 5𝑥 + 30𝑥 + 40𝑥 = 6𝑓 𝑡 − 20𝑔(𝑡) Solution Using the derivative property with zero initial conditions 5𝑠 𝑋 𝑠 + 30𝑠𝑋 𝑠 + 40𝑋 𝑠 = 6𝐹 𝑠 − 20𝐺(𝑠) Solve for 𝑋(𝑠) 20 𝑋 𝑠 = 𝐹 𝑠 − 𝐺(𝑠) 5𝑠 + 30𝑠 + 40 5𝑠 + 30𝑠 + 40 The transfer function for a specific input can be obtained by temporarily setting the other inputs equal to zero 𝑋(𝑠) 𝑋(𝑠) 20 = , = 𝐹(𝑠) 5𝑠 + 30𝑠 + 40 𝐺(𝑠) 5𝑠 + 30𝑠 + 40 §4.Transfer Function - Example 3.4.2 A System of Equations a.Obtain the transfer functions 𝑋(𝑠)/𝑉(𝑠) and 𝑌(𝑠)/𝑉(𝑠) of the following system of equations 𝑥 = −3𝑥 + 2𝑦 𝑦 = −9𝑦 − 4𝑥 + 3𝑣(𝑡) b.Obtain the forced response for 𝑥(𝑡) and 𝑦(𝑡) if the input is 𝑣(𝑡) = 5𝑢𝑠 (𝑡) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.65 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.66 §4.Transfer Function Solution a.Transform the system equations with zero initial conditions 𝑠𝑋 𝑠 = −3𝑋 𝑠 + 2𝑌 𝑠 𝑠𝑌 𝑠 = −9𝑌 𝑠 − 4𝑋 𝑠 + 3𝑉(𝑠) 𝑠+3 (1) ⟹𝑌 𝑠 = 𝑋(𝑠) 𝑠+3 𝑠+3 ⟹𝑠 𝑋 𝑠 = −9 𝑋 𝑠 − 4𝑋 𝑠 + 3𝑉(𝑠) 2 Then solve for 𝑋(𝑠)/𝑉(𝑠) to obtain 𝑋(𝑠) (2) = 𝑉(𝑠) 𝑠 + 12𝑠 + 35 Now substitute eq.(2) into eq.(1) to obtain 𝑌(𝑠) 𝑠 + 𝑋(𝑠) 𝑠 + 3(𝑠 + 3) = = = (3) 𝑉(𝑠) 𝑉(𝑠) 𝑠2 + 12𝑠 + 35 𝑠2 + 12𝑠 + 35 §4.Transfer Function b.From eq.(2) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Fluid and Thermal Systems 6 30 𝑉𝑠 = = 𝑠2 + 12𝑠 + 35 𝑠 + 12𝑠 + 35 𝑠 𝑠(𝑠 + 5)(𝑠 + 7) The partial-fraction expansion 𝐶1 𝐶2 𝐶3 𝑋 𝑠 = + + 𝑠 𝑠+5 𝑠+7 where 𝐶1 = 6/7, 𝐶2 = −3, 𝐶3 = 15/7 The forced response 15 (4) 𝑥 𝑡 = − 3𝑒 −5𝑡 − 𝑒 −7𝑡 7 From eq.(1) 𝑦 = (𝑥 + 3𝑥)/2 From eq.(4), we obtain 30 𝑦 𝑡 = + 3𝑒 −5𝑡 − 𝑒 −7𝑡 7 𝑋𝑠 = Nguyen Tan Tien 11 8/25/2013 System Dynamics 3.67 Fluid and Thermal Systems System Dynamics 3.68 Fluid and Thermal Systems §5.Partial Fraction Expansion Use for the Inverse Laplace transform Most transforms occur in the form of a ratio of two polynomials 𝑁(𝑠) 𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 + ⋯ + 𝑏1 𝑠 + 𝑏0 𝑋 𝑠 = = 𝐷(𝑠) 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + ⋯ + 𝑎1 𝑠 + 𝑎0 with 𝑚 ≤ 𝑛, the method of partial fraction expansion can be used 1.Distinct Root Case If all the roots are distinct, 𝑋(𝑠) can be factored as follows 𝑁(𝑠) 𝑋𝑠 = , 𝑠 = −𝑟1,−𝑟2, … , −𝑟𝑛 𝑠 + 𝑟1 𝑠 + 𝑟2 … (𝑠 + 𝑟𝑛) 𝐶1 𝐶2 𝐶𝑛 = + +⋯ , 𝐶𝑖 = lim [𝑋(𝑠)(𝑠 + 𝑟𝑖 )] 𝑠→−𝑟𝑖 𝑠 + 𝑟1 𝑠 + 𝑟2 𝑠 + 𝑟𝑛 Each factor corresponds to an exponential function of time 𝑥 𝑡 = 𝐶1 𝑒 −𝑟1 𝑡 + 𝐶2 𝑒 −𝑟2 𝑡 + ⋯ + 𝐶𝑛 𝑒 −𝑟𝑛 𝑡 §5.Partial Fraction Expansion - Example 3.5.1 One Zero Root and One Negative Root Obtain the inverse Laplace transform of 𝑋 𝑠 = 𝑠(𝑠 + 3) Solution The partial-fraction expansion has the form 𝐶1 𝐶2 𝑋 𝑠 = = + 𝑠(𝑠 + 3) 𝑠 𝑠+3 Using the coefficient formula 5 𝐶1 = lim 𝑠 = lim = 𝑠→0 𝑠→0 𝑠 + 𝑠 𝑠+3 5 𝐶2 = lim (𝑠 + 3) = lim = − 𝑠→−3 𝑠→−3 𝑠 𝑠 𝑠+3 The inverse transform: 𝑥 𝑡 = 𝐶1 + 𝐶2 𝑒 −3𝑡 = (5/3) − (5/3)𝑒 −3𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.69 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.70 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion - Example 3.5.2 A Third-Order Equation Use two methods to obtain the solution of the following problem 𝑑3𝑥 𝑑2𝑥 𝑑𝑥 10 + 100 + 310 + 300𝑥 = 750𝑢𝑠 𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑥 = 2, 𝑥 = 4, 𝑥 = Solution a.Using the Laplace transform method 10 𝑠 3𝑋 𝑠 − 𝑥 − 𝑠𝑥 − 𝑠 𝑥 +100[𝑠 2𝑋 𝑠 − 𝑥 − 𝑠𝑥(0)] 750 +310 𝑠𝑋 𝑠 − 𝑥 + 300𝑋 𝑠 = 𝑠 Solving for 𝑋(𝑠) using the given initial values to obtain 2𝑠3 + 24𝑠2 + 105𝑠 + 75 2𝑠3 + 24𝑠2 + 105𝑠 + 75 𝑋𝑠 = = 𝑠(𝑠3 + 10𝑠2 + 31𝑠 + 30) 𝑠(𝑠 + 2)(𝑠 + 3)(𝑠 + 5) §5.Partial Fraction Expansion The partial-fraction expansion 𝐶1 𝐶2 𝐶3 𝐶4 𝑋 𝑠 = + + + 𝑠 𝑠+2 𝑠+3 𝑠+5 The coefficients can be obtained from the formula 𝐶1 = lim 𝑠𝑋(𝑠) = 𝑠→0 55 𝐶2 = lim (𝑠 + 2)𝑋(𝑠) = 𝑠→−2 𝐶3 = lim (𝑠 + 3)𝑋(𝑠) = −13 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.71 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion 55 −2𝑡 10 + 𝑒 − 13𝑒 −3𝑡 + 𝑒 −5𝑡 and 𝑒 −5𝑡 die out faster than 𝑒 −2𝑡 10 the answer is 55 10 𝑥 𝑡 = + 𝑒 −2𝑡 − 13𝑒 −3𝑡 + 𝑒 −5𝑡 System Dynamics 3.72 (1) Nguyen Tan Tien Fluid and Thermal Systems 𝑡→∞ • 𝑡 > 4/3: the response is approximately 𝑥 𝑡 = + 55 −2𝑡 𝑒 • 𝑡 > 2: the response is approximately constant at 𝑥 = 5/2 The “hump” in the response is produced by the positive values of 𝑥(0) and 𝑥(0) HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑠→−5 §5.Partial Fraction Expansion b.The characteristic equation 10𝑠 + 100𝑠2 + 310𝑠 + 300 = 0, which has the roots 𝑠 = −2, −3, −5 can be obtained from ODE The characteristic equation roots generate the solution 𝑥 𝑡 = 𝐶1 + 𝐶2 𝑒 −2𝑡 + 𝐶3 𝑒 −3𝑡 + 𝐶4 𝑒 −5𝑡 (2) The solution will approach zero as 𝑡 → ∞, leaving only a constant term produced by the step input 𝐶1 = lim 𝑥(𝑡) 𝑥 𝑡 = The terms 𝑒 −3𝑡 𝑠→−3 𝐶4 = lim (𝑠 + 5)𝑋(𝑠) = Nguyen Tan Tien 𝐶1 can be found by setting the derivatives equal to zero in the ODE and solving for 𝑥 10𝑥 + 100𝑥 + 310𝑥 + 300𝑥 = 750𝑢𝑠 𝑡 750 ⟹ 𝐶1 = lim 𝑥(𝑡) = = 𝑡→∞ 300 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 12 8/25/2013 System Dynamics 3.73 Fluid and Thermal Systems System Dynamics 3.74 Fluid and Thermal Systems §5.Partial Fraction Expansion The values of the remaining coefficients can be found with the initial conditions by differentiating (2) to obtain 𝑥 = 𝐶1 + 𝐶2 + 𝐶3 + 𝐶4 = + 𝐶2 + 𝐶3 + 𝐶4 =2 𝑥 = −2𝐶2 − 3𝐶3 − 5𝐶4 = 𝑥 = 4𝐶2 + 9𝐶3 + 25𝐶4 = The solution is 𝐶1 = 55/6, 𝐶2 = −13, and 𝐶3 = 10/3 And the answer is 55 10 𝑥 𝑡 = + 𝑒 −2𝑡 − 13𝑒 −3𝑡 + 𝑒 −5𝑡 §5.Partial Fraction Expansion 2.Repeated Root Case Suppose that 𝑝 of the roots have the same value 𝑠 = −𝑟1, and the remaining (𝑛 − 𝑝) roots are distinct and real Then 𝑋(𝑠) is of the form 𝑁(𝑠) 𝑋 𝑠 = 𝑠 + 𝑟1 𝑝 𝑠 + 𝑟𝑝+1 𝑠 + 𝑟𝑝+2 … (𝑠 + 𝑟𝑛 ) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.75 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion The coefficients for the repeated roots are found from 𝐶1 = lim 𝑋 𝑠 𝑠 + 𝑟1 𝑝 𝑠→−𝑟1 𝐶2 = lim 𝑠→−𝑟1 𝑑 [𝑋 𝑠 𝑠 + 𝑟1 𝑝 ] 𝑑𝑠 ⋮ 𝐶𝑖 = lim 𝑠→−𝑟1 𝑑𝑖−1 [𝑋 𝑠 𝑠 + 𝑟1 𝑝 ] , 𝑖 = 1,2, … , 𝑝 𝑖 − ! 𝑑𝑠 𝑖−1 The solution for the time function is 𝑡 𝑝−1 −𝑟 𝑡 𝑡 𝑝−2 𝑓 𝑡 = 𝐶1 𝑒 + 𝐶2 𝑒 −𝑟1 𝑡 + ⋯ 𝑝−1 ! 𝑝−2 ! +𝐶𝑝 𝑒 −𝑟1 𝑡 + ⋯ + 𝐶𝑝+1 𝑒 −𝑟𝑝+1 𝑡 + ⋯ + 𝐶𝑛 𝑒 −𝑟𝑛 𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.77 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion 5/3 𝐶2 + 𝐶3 𝑠 + 𝐶1 + 4𝐶2 𝑠 + 4𝐶1 = 𝑠 (𝑠 + 4) 𝑠 (𝑠 + 4) Comparing numerators to obtain 𝐶2 + 𝐶3 = 𝐶1 = 5/12 𝐶1 + 4𝐶2 = ⟹ 𝐶2 = −5/48 4𝐶1 = 5/3 𝐶3 = 5/48 𝑋 𝑠 = System Dynamics 𝐶1 𝑠 + 𝑟1 𝑝 𝐶2 +⋯ 𝑠 + 𝑟1 𝑝−1 𝐶𝑝 𝐶𝑝+1 𝐶𝑛 + + ⋯+ +⋯+ 𝑠 + 𝑟1 𝑠 + 𝑟𝑝+1 𝑠 + 𝑟𝑛 + 3.76 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion - Example 3.5.3 One Negative Root and Two Zero Roots Compare two methods for obtaining the inverse Laplace transform of 𝑋 𝑠 = 𝑠 (3𝑠 + 12) Solution The partial-fraction expansion has the form 5 𝐶1 𝐶2 𝐶3 𝑋 𝑠 = = = + + 𝑠 (3𝑠 + 12) 𝑠 (𝑠 + 4) 𝑠 𝑠 𝑠 + Using the coefficient formulas 5 𝐶1 = lim 𝑠 2 = lim = 𝑠→0 𝑠→0 3(𝑠 + 4) 3𝑠 (𝑠 + 4) 12 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.78 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion 𝑑 𝑑 𝐶2 = lim 𝑠 = lim 𝑠→0 𝑑𝑠 𝑠→0 𝑑𝑠 3(𝑠 + 4) 3𝑠 (𝑠 + 4) 5 = lim − =− 𝑠→0 (𝑠 + 4)2 48 5 𝐶3 = lim (𝑠 + 4) = lim = 𝑠→−4 𝑠→−4 3𝑠 3𝑠 (𝑠 + 4) 48 The inverse transform is 5 𝑡− + 𝑒 −4𝑡 12 48 48 For this example the two methods are roughly equivalent • For repeated roots, coefficient formula requires that we obtain the derivative of a ratio of functions • The LCD method requires three equations to be solved for three unknowns 𝑥 𝑡 = 𝐶1 𝑡 + 𝐶2 + 𝐶3 𝑒 −4𝑡 = HCM City Univ of Technology, Faculty of Mechanical Engineering The expansion is Nguyen Tan Tien With the LCD (least common denominator) method we have 5/3 𝐶1 𝐶2 𝐶3 = + + 𝑠 (𝑠 + 4) 𝑠 𝑠 𝑠 + 𝐶1 𝑠 + + 𝐶2 𝑠 𝑠 + + 𝐶3 𝑠 = 𝑠 (𝑠 + 4) 𝐶1 + 𝐶3 𝑠 + 𝐶1 + 4𝐶2 𝑠 + 4𝐶1 = 𝑠 (𝑠 + 4) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 13 8/25/2013 System Dynamics 3.79 Fluid and Thermal Systems System Dynamics 3.80 Fluid and Thermal Systems §5.Partial Fraction Expansion - Example 3.5.4 Ramp Response of a First-Order Model Use the Laplace transform to solve the following problem 3𝑥 + 12𝑥 = 5𝑡 𝑥 =0 Solution Taking the transform of both sides of the equation 𝑠𝑋 𝑠 − 𝑥 + 12𝑋 𝑠 = 𝑠 Solve for 𝑋(𝑠) using the given value of 𝑥(0) 5 5 𝑋 𝑠 = = − + 𝑠 (3𝑠 + 12) 12 𝑠 48 𝑠 48 𝑠 + The inverse transform 5 𝑥 𝑡 = 𝑡− + 𝑒 −4𝑡 12 48 48 §5.Partial Fraction Expansion - Example 3.5.5 Two Repeated Roots and One Distinct Root Obtain the inverse Laplace transform of 𝑋 𝑠 = 𝑠 + (𝑠 + 5) Solution The partial-fraction expansion has the form 𝐶1 𝐶2 𝐶3 𝑋 𝑠 = = + + 𝑠 + (𝑠 + 5) 𝑠+3 𝑠+3 𝑠+5 where 7 𝐶1 = lim 𝑠 + = lim = 𝑠→−3 𝑠→−3 𝑠 + 𝑠 + (𝑠 + 5) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.81 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion 7 = lim = 𝑠→−3 𝑠 + 𝑠 + (𝑠 + 5) 𝑑 𝑑 𝐶2 = lim 𝑠+3 = lim 𝑠→−3 𝑑𝑠 𝑠→−3 𝑑𝑠 𝑠 + 𝑠 + (𝑠 + 5) −7 = lim =− 𝑠→−3 (𝑠 + 5)2 7 𝐶3 = lim 𝑠 + = lim = 𝑠→−5 𝑠→−5 (𝑠 + 3)2 𝑠 + (𝑠 + 5) 𝐶1 = lim 𝑠→−3 𝑠+3 The inverse transform 𝑥 𝑡 = 𝐶1 𝑡𝑒 −3𝑡 + 𝐶2 𝑒 −3𝑡 + 𝐶3 𝑒 −5𝑡 7 = 𝑡𝑒 −3𝑡 − 𝑒 −3𝑡 + 𝑒 −5𝑡 4 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.83 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion 7 𝑥 𝑡 = 𝑡𝑒 −3𝑡 − 𝑒 −3𝑡 + 𝑒 −5𝑡 4 The “hump” in the response is caused by the multiplicative factor of 𝑡 in the input (7/2)𝑡𝑒 −3𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics 3.82 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion - Example 3.5.6 Exponential Response of a First-Order Model Use the Laplace transform to solve the following problem 𝑥 + 5𝑥 = 7𝑡𝑒 −3𝑡 , 𝑥 =0 Solution Taking the transform of both sides of the equation 𝑠𝑋 𝑠 − 𝑥 + 5𝑋 𝑠 = (𝑠 + 3)2 Solve for 𝑋(𝑠) using the given value of 𝑥(0) 7 7 𝑋 𝑠 = = − + (𝑠 + 3)2 (𝑠 + 5) 2(𝑠 + 3)2 𝑠 + 4(𝑠 + 5) The inverse transform 7 𝑥 𝑡 = 𝑡𝑒 −3𝑡 − 𝑒 −3𝑡 + 𝑒 −5𝑡 4 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.84 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion - Example 3.5.7 Four Repeated Roots Choose the most convenient method for obtaining the inverse transform of 𝑠2 + 𝑋 𝑠 = 𝑠 (𝑠 + 1) Solution There are four repeated roots (𝑠 = 0) and one distinct root, so the expansion is 𝐶1 𝐶2 𝐶3 𝐶4 𝐶5 𝑋 𝑠 = 4+ 3+ 2+ + 𝑠 𝑠 𝑠 𝑠 𝑠+1 There are four repeated roots, to find the coefficients would require taking the first, second, and third derivatives of the ratio (𝑠 + 2)/(𝑠 + 1) Therefore the LCD method is easier to use for this problem HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 14 8/25/2013 System Dynamics 3.85 Fluid and Thermal Systems System Dynamics 3.86 Fluid and Thermal Systems §5.Partial Fraction Expansion Using the LCD to obtain 𝐶1 𝑠 + + 𝐶2𝑠 𝑠 + + 𝐶3𝑠2 𝑠 + + 𝐶4𝑠3 𝑠 + + 𝐶5𝑠4 𝑋 𝑠 = 𝑠4(𝑠 + 1) 𝐶5 + 𝐶4 𝑠4 + 𝐶4 + 𝐶3 𝑠3 + 𝐶3 + 𝐶2 𝑠2 + 𝐶2 + 𝐶1 𝑠 + 𝐶1 = 𝑠4(𝑠 + 1) Comparing numerators 𝑠2 + = 𝐶5 + 𝐶4 𝑠4 + 𝐶4 + 𝐶3 𝑠3 + 𝐶3 + 𝐶2 𝑠2 + 𝐶2 + 𝐶1 𝑠 + 𝐶1 ⟹ 𝐶1 = 2, 𝐶2 + 𝐶1 = 0, 𝐶3 + 𝐶2 = 1, 𝐶4 + 𝐶3 = 0, 𝐶5 + 𝐶4 = ⟹ 𝐶1 = 2, 𝐶2 = −2, 𝐶3 = 3, 𝐶4 = −3, 𝐶5 = So the expansion 2 3 𝑋 𝑠 = 4− 3+ 2− + 𝑠 𝑠 𝑠 𝑠 𝑠+1 ⟹ 𝑥 𝑡 = 𝑡 − 𝑡 + 3𝑡 − + 3𝑒 −𝑡 §5.Partial Fraction Expansion 3.Complex Root - Example 3.5.8 Two Complex Roots Use two methods to obtain the inverse Laplace transform of 𝑋 𝑠 = (3𝑠 + 7)/(4𝑠 + 24𝑠 + 136) Solution a.Rewrite the equation in the form 3𝑠 + 3𝑠 + 𝑋𝑠 = = 4𝑠 + 24𝑠 + 136 (𝑠 + 3)2+52 𝑠+3 𝐶2𝑠 + (5𝐶1 + 3𝐶2) = 𝐶1 + 𝐶2 = (𝑠 + 3)2+52 (𝑠 + 3)2+52 (𝑠 + 3)2+52 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.87 Nguyen Tan Tien Fluid and Thermal Systems Comparing the numerators to obtain 𝐶1 = −2/5, 𝐶2 = The inverse transform is 1 𝑥 𝑡 = 𝐶1𝑒−3𝑡𝑠𝑖𝑛5𝑡 + 𝐶2𝑒−3𝑡𝑐𝑜𝑠5𝑡 = − 𝑒−3𝑡𝑠𝑖𝑛5𝑡 + 𝑒−3𝑡𝑐𝑜𝑠5𝑡 4 10 System Dynamics 3.88 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion b.The denominator roots are distinct and the expansion gives 3𝑠+7 3𝑠+7 𝐶1 𝐶2 𝑋𝑠 = = = + 4𝑠 +24𝑠+136 4(𝑠+3−5𝑗)(𝑠+3+5𝑗) 𝑠+3−5𝑗 𝑠+3+5𝑗 3𝑠+ −2+ 15𝑗 15+2𝑗 𝐶1 = lim (𝑠 +3− 5𝑗)𝑋(𝑠) = lim = = 𝑠→−3+5𝑗 𝑠→−3+5𝑗 4(𝑠+ 3+5𝑗) 40𝑗 40 This can be expressed in complex exponential form as follows 15 + 2𝑗 𝑗𝜙 229 𝑗𝜙 𝐶1 = 𝐶1 𝑒𝑗𝜙 = 𝑒 = 𝑒 ,𝜙 = 𝑡𝑎𝑛−1 = 0.1326𝑟𝑎𝑑 40 40 15 3𝑠 + + 15𝑗 15− 2𝑗 𝐶2 = lim (𝑠 + + 5𝑗)𝑋(𝑠) = lim = = 𝑠→−3−5𝑗 𝑠→−3−5𝑗 4(𝑠 + − 5𝑗) 40𝑗 40 Note that 𝐶1 and 𝐶2 are complex conjugates This will always be the case for coefficients of complex-conjugate roots in a partial-fraction expansion Thus 𝐶2 = 𝐶1 𝑒 −𝑗𝜙 = 229𝑒 −0.1326𝑗 /40 §5.Partial Fraction Expansion The inverse transform gives 𝑥 𝑡 = 𝐶1 𝑒 (−3+5𝑗)𝑡 + 𝐶2 𝑒 (−3−5𝑗)𝑡 = 𝐶1 𝑒 −3𝑡 𝑒 5𝑗𝑡 + 𝐶2 𝑒 −3𝑡 𝑒 −5𝑗𝑡 = 𝐶1 𝑒 −3𝑡 𝑒 5𝑡+𝜙 𝑗 + 𝑒 − 5𝑡+𝜙 𝑗 = 𝐶1 𝑒 −3𝑡 𝑐𝑜𝑠(5𝑡 + 𝜙) where we have used the relation 𝑒𝑗𝜃 + 𝑒 −𝑗𝜃 = 2𝑐𝑜𝑠𝜃, which can be derived from the Euler identity Thus 229 −3𝑡 𝑥 𝑡 = 𝑒 𝑐𝑜𝑠(5𝑡 + 0.1326) 20 This answer is equivalent to that found in part (a), as can be seen by applying the trigonometric identity cos 5𝑡 + 𝜙 = 𝑐𝑜𝑠5𝑡𝑐𝑜𝑠𝜙 − 𝑠𝑖𝑛5𝑡𝑠𝑖𝑛𝜙 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.89 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.90 Nguyen Tan Tien Fluid and Thermal Systems §5.Partial Fraction Expansion - Example 3.5.9 Free Response of a 2nd-Order Model with Complex Roots Use the Laplace transform to solve the following problem 35 4𝑥 + 24𝑥 + 136𝑥 = 0, 𝑥 = ,𝑥 = − 4 Solution Taking the transform of both sides of the equation, we obtain 𝑠2𝑋 𝑠 − 𝑥 𝑠 − 𝑥 + 24 𝑠𝑋 𝑠 − 𝑥 + 136𝑋 𝑠 = Solve for 𝑋(𝑠) using the given values of 𝑥(0) and 𝑥(0) 𝑥 𝑠 + 𝑥 + 24𝑥(0) 3𝑠 + 𝑋 𝑠 = = 4𝑠 + 24𝑠 + 136 4(𝑠 + 6𝑠 + 34) 𝑠+3 =− + 10 𝑠 + + 52 (𝑠 + 3)2 +52 ⟹ 𝑓 𝑡 = − 𝑒 −3𝑡 𝑠𝑖𝑛5𝑡 + 𝑒 −3𝑡 𝑐𝑜𝑠5𝑡 10 §6.The Impulse and Numerator Dynamics - The effects of any inputs starting at 𝑡 = are not felt by the system until an infinitesimal time later, at 𝑡 = 0+ - The dependent variable 𝑥(𝑡) and its derivatives not change between 𝑡 = and 𝑡 = 0+, and thus the solution 𝑥(𝑡) obtained from the ODE will match the given initial conditions when 𝑥(𝑡) and its derivatives are evaluated at 𝑡 = - However, we will now investigate the behavior of some systems for which 𝑥(0) = 𝑥(0+), or 𝑥(0) = 𝑥(0+), and so forth for higher derivatives - The initial value theorem gives the value at 𝑡 = 0+, which for some models is not necessarily equal to the value at 𝑡 = In these cases the solution of the differential equation is correct only for 𝑡 > This phenomenon occurs in models having impulse inputs and in models containing derivatives of a discontinuous input, such as a step function HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 15 8/25/2013 System Dynamics 3.91 Fluid and Thermal Systems System Dynamics 3.92 Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics 1.The Impulse - Impulse: an input that is suddenly applied and removed after a very short time - Rectangular pulse: a constant input that is suddenly removed • its transform is 𝑀(1 − 𝑒 −𝑠𝐷 )/𝑠 • the strength of the pulse ≡ area under the pulse 𝐴 = 𝑀𝐷 If 𝐴 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 and 𝐷 ⟶ 0: we obtain the impulse - The transform 𝐹(𝑠) of an impulse − 𝑒−𝑠𝐷 𝐴 − 𝑒−𝑠𝐷 𝐴𝑠𝑒−𝑠𝐷 𝐹 𝑠 = lim 𝑀 = lim = lim =𝐴 𝐷→0 𝐷→0 𝐷 𝐷→0 𝑠 𝑠 𝑠 If 𝐴 = 1: unit impulse or Dirac delta function 𝛿(𝑡) - Consider the impulse 𝛿(𝑡) to start at time 𝑡 = and finish at 𝑡 = 0+, with its effects first felt at 𝑡 = 0+ §6.The Impulse and Numerator Dynamics - Example 3.6.1 Impulse Response of a Simple First-Order Model Obtain the unit-impulse response of the model 𝑥 = 𝛿(𝑡) a by separation of variables b with the Laplace transform The initial condition is 𝑥(0) = , what is the value of 𝑥(0+) ? Solution a Integrate both sides of the equation 𝑥 = 𝛿(𝑡) to obtain HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.93 Nguyen Tan Tien Fluid and Thermal Systems 𝑥(𝑡) 𝑡 𝑥 𝑡 𝑑𝑥 = 𝑥(0) 𝛿 𝑡 𝑑𝑡 or 𝑥 𝑡 −𝑥 =1⟹𝑥 𝑡 =𝑥 +1=3+1=4 This is the solution for 𝑡 > but not for 𝑡 = Thus, 𝑥 0+ = but 𝑥 = 3, so the impulse has changed 𝑥(𝑡) instantaneously from to System Dynamics 3.94 Nguyen Tan Tien Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics b The transformed equation is 𝑠𝑋 𝑠 − 𝑥 = or + 𝑥(0) 𝑋 𝑠 = 𝑠 which give the solution −1 𝑥 𝑡 = ℒ 𝑋 𝑠 = [1 + 𝑥 ] × = Note that the initial value used with the derivative property is the value of 𝑥 at 𝑡 = The initial value theorem gives + 𝑥(0) 𝑥 0+ = lim 𝑠𝑋(𝑠) = lim 𝑠 =1+3 =4 𝑠→∞ 𝑠→∞ 𝑠 which is correct §6.The Impulse and Numerator Dynamics - Example 3.6.2 Impulse Response of a First-Order Model Obtain the unit-impulse response of the following model 𝑋(𝑠)/𝐹(𝑠) = 1/(𝑠 + 5) The initial condition is 𝑥(0) = What is the value of 𝑥(0+) ? Solution Because 𝑓(𝑡) = 𝛿(𝑡), 𝐹(𝑠) = 1, and the response is obtained from 1 𝑋 𝑠 = 𝐹 𝑠 = 𝑠+5 𝑠+5 The response is 𝑥(𝑡) = 𝑒 −5𝑡 for 𝑡 > This gives 𝑥 0+ = lim 𝑥(𝑡) = lim 𝑒 −5𝑡 = HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑡→0+ 𝑠→∞ System Dynamics 3.95 Nguyen Tan Tien Fluid and Thermal Systems 𝑡→0+ the impulse input has changed 𝑥 from 𝑥 = to 𝑥(0+) = This same result can be obtained from the initial value theorem 𝑥 0+ = lim 𝑠𝑋(𝑠) = lim [𝑠/(𝑠 + 5)] = System Dynamics 𝑠→∞ 3.96 Nguyen Tan Tien Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics - Example 3.6.3 Impulse Response of a Simple Second-Order Model Obtain the unit-impulse response of the model 𝑥 = 𝛿(𝑡) a by separation of variables b with the Laplace transform The initial condition is 𝑥(0) = and 𝑥 = 10, what is the value of 𝑥(0+) and 𝑥(0+) ? Solution a Let 𝑣(𝑡) ≡ 𝑥(𝑡) ⟶ 𝑣 = 𝛿(𝑡) Integrate 𝑣 = 𝛿(𝑡) to obtain 𝑣(𝑡) = 𝑣(0) + = 10 + = 11 Thus 𝑥 0+ = 11 ≠ 𝑥(0) Now integrate 𝑥 = 𝑣 = 11 to obtain 𝑥(𝑡) = 𝑥(0) + 11𝑡 = + 11𝑡 Thus, 𝑥 0+ = = 𝑥(0) So for this model the unit-impulse input changes 𝑥 from 𝑡 = to 𝑡 = 0+ but does not change 𝑥 §6.The Impulse and Numerator Dynamics b The transformed equation is 𝑠 𝑋 𝑠 − 𝑠𝑥 − 𝑥 = 𝑠𝑥 + 𝑥 + 5𝑠 + 11 11 ⟹𝑋 𝑠 = = = + 𝑠2 𝑠2 𝑠 𝑠 ⟹ 𝑥 𝑡 = ℒ −1 {𝑋 𝑠 } = + 11𝑡 and 𝑥 𝑡 = 11 Note that the initial values used with the derivative property are the values at 𝑡 = The initial value theorem gives 5𝑠 + 11 𝑥 0+ = lim 𝑠𝑋(𝑠) = lim 𝑠 =5 𝑠→∞ 𝑠→∞ 𝑠2 and because ℒ 𝑥 = 𝑠𝑋 𝑠 − 𝑥(0) 5𝑠 + 11 𝑥 0+ = lim 𝑠[𝑠𝑋 𝑠 − 𝑥 ] = lim 𝑠 − = 11 𝑠→∞ 𝑠→∞ 𝑠2 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien as we found in part (a) Nguyen Tan Tien 16 8/25/2013 System Dynamics 3.97 Fluid and Thermal Systems System Dynamics 3.98 Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics - Example 3.6.4 Impulse Response of a Second-Order Model Obtain the unit-impulse response of the following model The initial conditions are 𝑥(0) = 0, 𝑥(0) = What are the values of 𝑥(0+) and 𝑥(0+) ? 𝑋(𝑠) = 𝐹(𝑠) 2𝑠 + 14𝑠 + 20 Solution Because 𝐹(𝑠) = 1, the response is obtained from 𝑋 𝑠 = 𝐹 𝑠 2𝑠 + 14𝑠 + 20 1 1 = = − 2𝑠 + 14𝑠 + 20 𝑠 + 𝑠 + and the response is 𝑥 𝑡 = (𝑒 −2𝑡 − 𝑒 −5𝑡 )/6 §6.The Impulse and Numerator Dynamics This gives 𝑒 −2𝑡 − 𝑒 −5𝑡 𝑥 0+ = lim 𝑥 𝑡 = lim =0 𝑡→0+ 𝑡→0+ −2𝑒 −2𝑡 + 5𝑒 −5𝑡 𝑥 0+ = lim 𝑥(𝑡) = lim = 𝑡→0+ 𝑡→0+ HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.99 Nguyen Tan Tien Fluid and Thermal Systems So the impulse input has not changed 𝑥 between 𝑡 = and 𝑡 = 0+ but has changed 𝑥 from to 1/2 These results could have been obtained from the initial value theorem 𝑥 0+ = lim 𝑠𝑋(𝑠) = lim 𝑠 =0 𝑠→∞ 𝑠→∞ 2𝑠 + 14𝑠 + 20 1 𝑥 0+ = lim 𝑠[𝑠𝑋 𝑠 − 𝑥(0)] = lim 𝑠 = 𝑠→∞ 𝑠→∞ 2𝑠 + 14𝑠 + 20 System Dynamics 3.100 Nguyen Tan Tien Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics In summary, be aware that the solution 𝑥(𝑡) and its derivatives 𝑥(𝑡), 𝑥(𝑡), … will match the given initial conditions at 𝑡 = only if there are no impulse inputs and no derivatives of inputs that are discontinuous at 𝑡 = If 𝑋(𝑠) is a rational function • if the degree of the numerator of 𝑋(𝑠) is less than the degree of the denominator, then the initial value theorem will give a finite value for 𝑥(0+) • if the degrees are equal, then initial value is undefined and the initial value theorem is invalid, 𝑥(0+) is undefined The latter situation corresponds to an impulse in 𝑥(𝑡) at 𝑡 = and therefore 𝑥(0+) is undefined For example, consider the 9𝑠 + 23 𝑋 𝑠 = =9− ⟹ 𝑥 𝑡 = 9𝛿 𝑡 − 23𝑒 −3𝑡 𝑠+3 𝑠+3 therefore 𝑥(0+) is undefined §6.The Impulse and Numerator Dynamics 2.Numerator Dynamics The following model contains a derivative of the input 𝑔(𝑡) 5𝑥 + 10𝑥 = 2𝑔 𝑡 + 10𝑔(𝑡) Its transfer function is 𝑋(𝑠) 2𝑠 + 10 = 𝐺(𝑠) 5𝑠 + 10 Note that the input derivative 𝑔(𝑡) results in an 𝑠 term in the numerator of the transfer function, and such a model is said to have numerator dynamics With such models we must proceed carefully if the input is discontinuous, as is the case with the step function, because the input derivative produces an impulse when acting on a discontinuous input HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.101 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.102 Nguyen Tan Tien Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics The unit impulse 𝛿(𝑡) is the time-derivative of the unit-step function 𝑢𝑠 (𝑡); that is, 𝑑 𝛿 𝑡 = 𝑢𝑠 (𝑡) 𝑑𝑡 This result does not contradict common sense, because the step function changes from at 𝑡 = to at 𝑡 = 0+ in an infinitesimal amount of time Therefore its derivative should be infinite during this time To further indicate the correctness of this relation, we integrate both sides and note that the area under the unit impulse is unity Thus 0+ 0+ 𝑑 𝛿 𝑡 𝑑𝑡 = 𝑢𝑠 𝑡 𝑑𝑡 = 𝑢𝑠 0+ − 𝑢𝑠 = − = 0 𝑑𝑡 which gives = Thus an input derivative will create an impulse in response to a step input §6.The Impulse and Numerator Dynamics An input derivative will create an impulse in response to a step input For example, consider the model 5𝑥 + 10𝑥 = 2𝑔 𝑡 + 10𝑔(𝑡) If the input 𝑔(𝑡) = 𝑢𝑠 (𝑡), the model is equivalent to 5𝑥 + 10𝑥 = 2𝛿 𝑡 + 10𝑢𝑠 𝑡 which has an impulse input Numerator dynamics can significantly alter the response, and the Laplace transform is a convenient and powerful tool for analyzing models having numerator dynamics HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 17 8/25/2013 System Dynamics 3.103 Fluid and Thermal Systems System Dynamics 3.104 Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics - Example 3.6.5 A First-Order Model with Numerator Dynamics Obtain the transfer function and investigate the response of the following model in terms of the parameter 𝑎 The input 𝑔(𝑡) is a unit-step function 5𝑥 + 10𝑥 = 𝑎 𝑔 𝑡 + 10𝑔(𝑡) 𝑥 =0 Solution Transforming the equation with 𝑥(0) = and solving for the ratio 𝑋(𝑠)/𝐺(𝑠) gives the transfer function 𝑋(𝑠) 𝑎𝑠 + 10 = 𝐺(𝑠) 5𝑠 + 10 Note that the model has numerator dynamics if 𝑎 ≠ For a unit-step input, 𝐺(𝑠) = 1/𝑠 and 𝑎𝑠 + 10 𝑎−5 𝑋 𝑠 = = + ⟹𝑥 𝑡 𝑠(5𝑠 + 10) 𝑠 𝑠+2 HCM City Univ of Technology, 𝑎 − 5Faculty of Mechanical Engineering Nguyen Tan Tien −2𝑡 =1+ 𝑒 §6.The Impulse and Numerator Dynamics System Dynamics System Dynamics 3.105 Fluid and Thermal Systems 𝑥 𝑡 =1+ 𝑎 − −2𝑡 𝑒 From this solution: 𝑥(0+) = 𝑎/5, which is not equal to 𝑥(0) unless 𝑎 = 0(which corresponds to the absence of numerator dynamics) The initial condition is different for each case, but for all cases the response is essentially constant for 𝑡 > because of the term 𝑒 −2𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering 3.106 Nguyen Tan Tien Fluid and Thermal Systems §6.The Impulse and Numerator Dynamics - Example 3.6.6 A Second-Order Model with Numerator Dynamics Obtain the transfer function and investigate the response of the following model in terms of the parameter 𝑎 The input 𝑔(𝑡) is a unit-step function 3𝑥 + 18𝑥 + 24𝑥 = 𝑎𝑔 𝑡 + 6𝑔(𝑡) 𝑥 = 0, 𝑥 = Solution Transforming the equation with zero initial conditions and solving for the ratio 𝑋(𝑠)/𝐺(𝑠) gives the transfer function 𝑋(𝑠) 𝑎𝑠 + = 𝐺(𝑠) 3𝑠 + 18𝑠 + 24 Note that the model has numerator dynamics if 𝑎 ≠ For a unit-step input, 𝐺(𝑠) = 1/𝑠 and 𝑎𝑠 + 11 𝑎 − 3 − 2𝑎 𝑋 𝑠 = = + + 𝑠(3𝑠 + 18𝑠 + 24) 𝑠 𝑠+2 12 𝑠 + §6.The Impulse and Numerator Dynamics 𝑎 − −2𝑡 − 2𝑎 −4𝑡 The response is 𝑥 𝑡 = + 𝑒 + 𝑒 12 From this solution: 𝑥 0+ = = 𝑥(0), and 𝑥 0+ = 𝑎/3 ≠ 𝑥(0) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 3.107 Fluid and Thermal Systems §7.Additional Examples - Example 3.7.1 Transform of 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜑) a Derive the Laplace transform of the function 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜑) b Generalize the answer from part (a) to find the Laplace transform of 𝐴𝑒 −𝑎𝑡 𝑠𝑖𝑛(𝜔𝑡 + 𝜑) Solution a From the trigonometric identity 𝑠𝑖𝑛 𝜔𝑡 + 𝜙 = 𝑠𝑖𝑛𝜔𝑡𝑐𝑜𝑠𝜙 + 𝑐𝑜𝑠𝜔𝑡𝑠𝑖𝑛𝜙 ∞ ∞ 𝑠𝑖𝑛𝜔𝑡𝑐𝑜𝑠𝜙𝑒−𝑠𝑡𝑑𝑡 + 𝐴 ⟹ ℒ 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙 = 𝐴 𝑐𝑜𝑠𝜔𝑡𝑠𝑖𝑛𝜙𝑒−𝑠𝑡𝑑𝑡 = 𝐴𝑐𝑜𝑠𝜙ℒ 𝑠𝑖𝑛𝜔𝑡 + 𝐴𝑠𝑖𝑛𝜙ℒ 𝑠𝑖𝑛𝜔𝑡 𝜔 𝑠 = 𝐴𝑐𝑜𝑠𝜙 + 𝐴𝑠𝑖𝑛𝜙 𝑠 + 𝜔2 𝑠 + 𝜔2 𝑠𝑠𝑖𝑛𝜙 + 𝜔𝑐𝑜𝑠𝜙 ⟹ ℒ 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙 = 𝐴 𝑠2 + 𝜔2 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Notice that a “hump” in the response (called an “overshoot”) does not occur for smaller values of 𝑎 and the height of the hump increases as a increases However, the value of 𝑎 does not affect the steady-state response System Dynamics 3.108 Nguyen Tan Tien Fluid and Thermal Systems §7.Additional Examples b Following the same procedure and using the fact that 𝜔 ℒ 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 = (𝑠 + 𝑎)2 +𝜔 𝑠+𝑎 ℒ 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 = (𝑠 + 𝑎)2 +𝜔 ⟹ ℒ 𝐴𝑒 −𝑎𝑡 𝑠𝑖𝑛 𝜔𝑡 + 𝜙 = 𝐴𝑐𝑜𝑠𝜙ℒ 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 +𝐴𝑠𝑖𝑛𝜙ℒ 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 𝜔 = 𝐴𝑐𝑜𝑠𝜙 (𝑠 + 𝑎)2 +𝜔 𝑠+𝑎 +𝐴𝑠𝑖𝑛𝜙 (𝑠 + 𝑎)2 +𝜔 𝑠𝑠𝑖𝑛𝜙 + 𝑎𝑠𝑖𝑛𝜙 + 𝜔𝑐𝑜𝑠𝜙 ⟹ ℒ 𝐴𝑒 −𝑎𝑡 𝑠𝑖𝑛 𝜔𝑡 + 𝜙 = 𝐴 (𝑠 + 𝑎)2 +𝜔 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 18 8/25/2013 System Dynamics 3.109 Fluid and Thermal Systems §7.Additional Examples - Example 3.7.2 Response in the Form 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜑) 3𝑠 + 10 𝑠 + 16 obtain 𝑓(𝑡) in the form 𝑓(𝑡) = 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜑), where 𝐴 > Solution 𝑠𝑠𝑖𝑛𝜙 + 𝜔𝑐𝑜𝑠𝜙 ℒ 𝐴𝑠𝑖𝑛 𝜔𝑡 + 𝜙 = 𝐴 𝑠2 + 𝜔2 Comparing this with 𝐹(𝑠) to get 𝑠𝑖𝑛𝜙 = 3/𝐴, 𝑐𝑜𝑠𝜙 = 5/(2𝐴) 𝑠𝑖𝑛𝜙 3/𝐴 𝜙 = 𝑡𝑎𝑛−1 = 𝑡𝑎𝑛−1 = 𝑡𝑎𝑛−1 − 0.876𝑟𝑎𝑑 𝑐𝑜𝑠𝜙 10/4𝐴 10/4 2 10 𝑠𝑖𝑛2 𝜙 + 𝑐𝑜𝑠 = + = ⟹ 𝐴 = 61/2 𝐴 4𝐴 The solution is 𝑓 𝑡 = 0.5 61𝑠𝑖𝑛(4𝑡 + 0.876) Given that 𝐹 𝑠 = HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.111 Nguyen Tan Tien Fluid and Thermal Systems §7.Additional Examples This can be expressed as a single fraction as follows 𝐶1 (𝑠 + 2)2 +25 + 𝐶2 𝑠(𝑠𝑠𝑖𝑛𝜙 + 2𝑠𝑖𝑛𝜙 + 5𝑐𝑜𝑠𝜙) 𝑋 𝑠 = 𝑠 (𝑠 + 2)2 +25 𝐶1 + 𝐶2𝑠𝑖𝑛𝜙 𝑠2 + 4𝐶1 + 2𝐶2𝑠𝑖𝑛𝜙 + 5𝐶2𝑐𝑜𝑠𝜙 𝑠 + 29𝐶1 = (2) 𝑠 (𝑠 + 2)2+25 Comparing the numerators of eq.s (1) and (2) 𝐶1 + 𝐶2 𝑠𝑖𝑛𝜙 = 𝐶 = 0.0575 4𝐶1 + 2𝐶2𝑠𝑖𝑛𝜙 + 5𝐶2 𝑐𝑜𝑠𝜙 = 15 ⟹ 𝐶2 = 2.918 29𝐶1 = 5/3 Thus the solution is 𝑥 𝑡 = 0.0575 + 2.918𝑒 −2𝑡 𝑠𝑖𝑛(5𝑡 + 0.729) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.113 Nguyen Tan Tien Fluid and Thermal Systems §7.Additional Examples From the shifting property 𝑥 𝑡 = 𝑦 𝑡 𝑢𝑠 𝑡 − 𝑦 𝑡 − 𝑢𝑠 (𝑡 − 2) (1) To find 𝑦(𝑡), note that the denominator roots of 𝑌(𝑠) are 𝑠 = and 𝑠 = −4 Thus we can express 𝑌(𝑠) as follows 𝐶1 𝐶2 1 1 𝑌 𝑠 = + = − ⟹ 𝑦 𝑡 = − 𝑒 −4𝑡 𝑠 𝑠+4 𝑠 𝑠+4 4 and from eq.(1) 1 −4𝑡 1 𝑥 𝑡 = − 𝑒 − − 𝑒 −4 𝑡−2 𝑢𝑠 (𝑡 − 2) 4 4 1 So for ≤ 𝑡 ≤ 𝑥 𝑡 = − 𝑒 −4𝑡 4 1 1 for 𝑡>2 𝑥 𝑡 = − 𝑒 −4𝑡 − + 𝑒 −4 𝑡−2 4 4 = 0.25 𝑒 −4(𝑡−2) − 𝑒 −4𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.110 The roots are 𝑠 = 0, 𝑠 = −2 ± 5𝑗, the form of the solution is 𝑥 𝑡 = 𝐶1 + 𝐶2 𝑒 −2𝑡 𝑠𝑖𝑛(5𝑡 + 𝜙) 𝑠𝑠𝑖𝑛𝜙 + 2𝑠𝑖𝑛𝜙 + 5𝑐𝑜𝑠𝜙 ⟹ 𝑋 𝑠 = 𝐶1 + 𝐶2 𝑠 (𝑠 + 2)2 +25 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.112 Nguyen Tan Tien Fluid and Thermal Systems §7.Additional Examples - Example 3.7.4 Pulse Response of a First-Order Model Suppose a rectangular pulse 𝑃(𝑡) of unit height and duration is applied to the first-order model 𝑥 + 4𝑥 = 𝑃(𝑡), 𝑥 = Use the Laplace transform to determine the response Solution Taking the transform of both sides of the equation and noting that the initial condition is zero, we obtain − 𝑒 −2𝑠 𝑠𝑋 𝑠 + 4𝑋 𝑠 = 𝑃 𝑠 = 𝑠 Solve for 𝑋(𝑠) 𝑃(𝑠) − 𝑒 −2𝑠 𝑋 𝑠 = = 𝑠 + 𝑠(𝑠 + 4) Let 𝑌 𝑠 ≡ ⟹ 𝑋 𝑠 = − 𝑒 −2𝑠 𝑌 𝑠 𝑠(𝑠 + 4) HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.114 Nguyen Tan Tien Fluid and Thermal Systems §7.Additional Examples So for ≤ 𝑡 ≤ for Nguyen Tan Tien Fluid and Thermal Systems §7.Additional Examples - Example 3.7.3 Sine Form of the Response Obtain the solution to the following problem in the form of a sine function with a phase angle 3𝑥 + 12𝑥 + 87𝑥 = 5, 𝑥 = 2, 𝑥 = Solution Applying the Laplace transform we obtain 𝑠2𝑋 𝑠 − 𝑠𝑥 − 𝑥(0) + 12 𝑠𝑋 𝑠 − 𝑥(0) + 87𝑋 𝑠 = 5/𝑠 6𝑠 + 45𝑠 + 2𝑠 + 15𝑠 + 5/3 ⟹𝑋 𝑠 = = (1) 3𝑠(𝑠 + 4𝑠 + 29) 𝑠 (𝑠 + 2)2 +25 𝑡>2 1 −4𝑡 − 𝑒 4 1 −4𝑡 1 −4 𝑥 𝑡 = − 𝑒 − + 𝑒 4 4 −4(𝑡−2) = 0.25 𝑒 − 𝑒 −4𝑡 𝑥 𝑡 = HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑡−2 Nguyen Tan Tien 19 8/25/2013 System Dynamics 3.115 Fluid and Thermal Systems System Dynamics 3.116 Fluid and Thermal Systems §7.Additional Examples - Example 3.7.5 Series Solution Method Obtain an approximate, closed-form solution of the following problem for ≤ 𝑡 ≤ 0.5: 𝑥 + 𝑥 = 𝑡𝑎𝑛𝑡, 𝑥 = Solution Use separation of variables to solve this problem we obtain 𝑑𝑡 𝑥 + 𝑥 = 𝑡𝑎𝑛𝑡 ⟹ = 𝑑𝑡 𝑡𝑎𝑛𝑡 − 𝑥 so the variables not separate In general, when the input is a function of time, the equation 𝑥 + 𝑔(𝑥) = 𝑓(𝑡) does not separate The Laplace transform method cannot be used when the Laplace transform or inverse transform either does not exist or cannot be found easily In this example, the equation cannot be solved by the Laplace transform method, because the transform of 𝑡𝑎𝑛𝑡 does not exist §7.Additional Examples An approximate solution of the equation 𝑥 + 𝑥 = 𝑡𝑎𝑛𝑡 can be obtained by replacing 𝑡𝑎𝑛𝑡 with a series approximation 𝑡 2𝑡 17𝑡 𝜋 𝑡𝑎𝑛𝑡 = 𝑡 + + + +⋯ , 𝑡 < 15 315 Using the two-term series to solve the following problem 𝑡3 𝑥+𝑥 =𝑡+ , 𝑥 =0 Using the Laplace transform we obtain 1 3! 𝑠𝑋 𝑠 + 𝑋 𝑠 = + 𝑠 3𝑠 𝑠2 + 2 3 ⟹ 𝑋(𝑠) = = 4− 3+ 2− + 𝑠 (𝑠 + 1) 𝑠 𝑠 𝑠 𝑠 𝑠+1 ⟹ 𝑥 𝑡 = 𝑡 − 𝑡 + 3𝑡 − + 3𝑒 −𝑡 HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Nguyen Tan Tien 3.117 Fluid and Thermal Systems §8.Computing Expansion Coefficients with Matlab - Example 8.1 Real, Distinct Poles 6𝑠 + 57𝑠 + 120𝑠 + 80 𝑋 𝑠 = 𝑠 + 9𝑠 + 14 Solution Matlab [r,p,K] = residue([6,57,120,80],[1,9,14]) Result r = [5, 4], p = [−7, −2], K = [6, 3] 𝑋 𝑠 = 6𝑠 + + 9𝑠 + 38 1 = 𝑠+ +5 +4 𝑠 + 9𝑠 + 14 𝑘 𝑠 − (−7) 𝑟 𝑠 − (−2) 𝑘 𝑟 𝑝1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics Fluid and Thermal Systems §8.Computing Expansion Coefficients with Matlab - Example 8.3 Complex Poles 4𝑠 + 𝑋 𝑠 = 𝑠 + 6𝑠 + 34𝑠 Solution Matlab [r,p,K] = residue([4,1],[1,6,34,0]) Result r = [−0.0147 − 0.3912𝑖, −0.0147 + 0.3912𝑖, 0.0294] p = [−3 + 5𝑖, −3 − 5𝑖, 0] K=[] 𝑟1 𝑟2 𝑟3 −0.0147 − 0.3912𝑗 −0.0147 + 0.3912𝑗 0.0294 𝑋 𝑠 = + + 𝑠 − (−3 + 5𝑗) 𝑠 − (−3 − 5𝑗) 𝑠− 𝑝1 HCM City Univ of Technology, Faculty of Mechanical Engineering 𝑝2 𝑝3 Nguyen Tan Tien 3.118 Fluid and Thermal Systems §8.Computing Expansion Coefficients with Matlab - Example 8.2 Repeated Poles 12𝑠 + 𝑋 𝑠 = 𝑠 + 16𝑠 + 77𝑠 + 98 Solution Matlab [r,p,K] = residue([12,8],[1,16,77,98]) Result r =[0.64, 15.2, −0.64], p = [−7, −7, −2], K = [ ] 𝑋 𝑠 = + 0.64 𝑘 𝑝2 Nguyen Tan Tien 3.119 System Dynamics Nguyen Tan Tien 𝑟1 1 + 1.52 + (−0.64) 𝑠 − (−7) [𝑠 − (−7)]2 𝑠 − (−2) 𝑟 𝑟 𝑝1 HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.120 𝑝2 𝑝3 Nguyen Tan Tien Fluid and Thermal Systems §9.Transfer Function Analysis in Matlab 1.The tf and tfdata function tf creation of TF or conversion to TF tfdata quick access to TF data Example 𝑋(𝑠) = 𝐹(𝑠) 5𝑠 + 9𝑠 + Solution Matlab sys1 = tf(1, [5, 9, 4]) Example 𝑑3𝑥 𝑑2𝑥 𝑑𝑥 𝑑2𝑓 𝑑𝑓 +4 +7 + 3𝑥 = + + 2𝑓 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 Solution Matlab sys2 = tf([6, 9, 2],[5, 4, 7, 3]) HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien 20 8/25/2013 System Dynamics 3.121 Fluid and Thermal Systems System Dynamics 3.122 Fluid and Thermal Systems §9.Transfer Function Analysis in Matlab 2.ODE Solvers - The Control System Toolbox provides several solvers for linear models These solvers are categorized by the type of input function they can accept: some of these are a step input, an impulse input, and a general input function - Example 3.9.1 Step Response of a Second-Order Model 𝑋(𝑠) 𝑐𝑠 + Consider the model , plot = 𝐹(𝑠) 10𝑠 + 𝑐𝑠 + a The unit-step response for 𝑐 = using the time span selected by Matlab b The unit-step response for 𝑐 = over the range ≤ 𝑡 ≤ 15 c The unit-step responses for 𝑐 = and 𝑐 = over the range ≤ 𝑡 ≤ 15 Put the plots on the same graph d The step response for 𝑐 = 3, where the magnitude of the step input is 20 Use the time span selected by Matlab §9.Transfer Function Analysis in Matlab a The unit-step response for 𝑐 = using the time span selected by Matlab Matlab sys1 = tf([3,5],[10,3,5]); step(sys1) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.123 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.124 Nguyen Tan Tien Fluid and Thermal Systems §9.Transfer Function Analysis in Matlab b The unit-step response for 𝑐 = over the range ≤ 𝑡 ≤ 15 Matlab sys1 = tf([3,5],[10,3,5]); t = (0:0.01:15); step(sys1) §9.Transfer Function Analysis in Matlab c The unit-step responses for 𝑐 = and 𝑐 = over the range ≤ 𝑡 ≤ 15 Put the plots on the same graph Matlab sys1 = tf([3,5],[10,3,5]); sys2 = tf([8,5],[10,8,5]); t = (0:0.01:15); step(sys1,' ',sys2,' ',t) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering System Dynamics 3.125 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.126 Nguyen Tan Tien Fluid and Thermal Systems §9.Transfer Function Analysis in Matlab d The step response for 𝑐 = 3, where the magnitude of the step input is 20 Use the time span selected by Matlab Matlab sys1 = tf([3,5],[10,3,5]); [y, t] = step(sys1); plot(t,20*y),xlabel('t'),ylabel('x(t)') §9.Transfer Function Analysis in Matlab - Example 3.9.2 Impulse Response of Second-Order Models Obtained the response of the following model to a unit impulse 𝑋(𝑠) = 𝐹(𝑠) 2𝑠 + 14𝑠 + 20 Our analysis showed that if 𝑥(0) = 𝑥(0) = 0, then 𝑥 0+ = and 𝑥 0+ = 1/2 Use the impulse function to verify these results Solution Matlab sys1 = tf(1,[2, 14, 20]); impulse(sys1) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien 21 8/25/2013 System Dynamics 3.127 Fluid and Thermal Systems System Dynamics 3.128 §9.Transfer Function Analysis in Matlab From the plot we see that𝑥 0+ = and that 𝑥 0+ is positive as predicted We are unable to determine the exact value of 𝑥 0+ from this plot, so we multiply the transfer function by 𝑠 to obtain the transfer function for 𝑣 = 𝑥 §9.Transfer Function Analysis in Matlab 𝑉(𝑠) 𝑠 = 𝐹(𝑠) 2𝑠 + 14𝑠 + 20 Matlab sys2 = tf([1,0],[2, 14, 20]); impulse(sys1) HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Fluid and Thermal Systems 𝑥 0+ = 1/2 as predicted System Dynamics 3.129 Nguyen Tan Tien Fluid and Thermal Systems System Dynamics 3.130 §9.Transfer Function Analysis in Matlab - Example 3.9.3 Ramp Response with the lsim Function Plot the forced response of 𝑥 + 3𝑥 + 𝑥5 = 10𝑓(𝑡) to a ramp input, 𝑢(𝑡) = 1.5𝑡, over the time interval ≤ 𝑡 ≤ Solution Matlab t = linspace(0,2,300); f = 1.5*t; sys = tf(10, [1, 3, 5]); [y, t] = lsim(sys,f,t); plot(t,y,t,f),xlabel('t'),ylabel('x(t) and f(t)') §9.Transfer Function Analysis in Matlab The resulting plot HCM City Univ of Technology, Faculty of Mechanical Engineering HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Nguyen Tan Tien Fluid and Thermal Systems Nguyen Tan Tien 22 ... Solution Methods for Dynamic Models §2.Response Types and Stability System Dynamics 3.20 Solution Methods for Dynamic Models §2.Response Types and Stability The time constant is useful also for. .. appropriate trial -solution form