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ME 140L - Lecture 3 - Second Order Systems

# ME 140L - Lecture 3 - Second Order Systems - MECHATRONICS...

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1 MECHATRONICS LAB ME 140L SECOND-ORDER CIRCUIT ANALYSIS RECALL ME324: MECHANICAL SYSTEMS—2 nd ORDER SYSTEM An ideal mass-spring-damper system Constitutive equations: General Linear Energy Mass: p=p(V) p=MV 2 2 MV 2M 2 p = Damper: F=F(V) F=BV P=BV 2 Spring: F=F(x) F=kx 2 2 kx 2k 2 F = Compatibility/Continuity: m B m B x x x and x x x k k & & & = = = = M F x M k x M B x kx x B - F x M x m F : Law) s (Newton' Motion of Equation = + + = = & & & & & & & & The equation of motion for all second-order linear systems has the form: frequency atural n n ω ratio, amping d ξ able input vari u , interest of variable x : where M B and M k n 2 n 2 n n ω 2 ξ ω u x ω x ω 2 ξ x = = = = = = = + + & & &

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2 M F x M k x M B x = + + & & & By solving this ODE given F, one can characterize the motion of the system. Can change M, B, k to alter the motion of the system Special cases—consider: a. No damping—B=0 (no energy dissipation) M F x M k x = + & & What is frequency of oscillation? = M k n ω M F x 2 n ω x = + & & Note: solution is sinusoidal: t ω sin A x(t) n = b. No spring—k=0 ODE) order - st (1 M F v M B v x B x = + = + & & & & Note: solution is exponential: ) - e - A(1 v(t) α t = Unit step responses to the above two cases: Step Response Time (sec) Amplitude 0 5 10 15 20 25 30 35 40 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 1 2 3 4 5 6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Step Response Time (sec) Amplitude No damper No spring