ME 140L - Lecture 3 - Second Order Systems

ME 140L - Lecture 3 - Second Order Systems - 1 MECHATRONICS...

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Unformatted text preview: 1 MECHATRONICS LAB ME 140L SECOND-ORDER CIRCUIT ANALYSIS RECALL ME324: MECHANICAL SYSTEMS2 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 ratio, amping d able input vari u , interest of variable x : where M B and M k n 2 n 2 n n 2 u x x 2 x = = = = = = = + + & & & 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 casesconsider: a. No dampingB=0 (no energy dissipation) M F x M k x = + & & What is frequency of oscillation? = M k n M F x 2 n x = + & & Note: solution is sinusoidal: t sin A x(t) n = b. No springk=0 ODE) order - st (1 M F v M B v x B x = + = + & & & & Note: solution is exponential: )- e- A(1 v(t) t = Unit step responses to the above two cases: Step Response Time (sec) Amplitude 5 10 15 20 25 30 35 40 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1 2 3 4 5 6 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 3 I. CIRCUIT ANALYSIS: Second-Order Systems Called second-order system because the system response is described by one second-order differential equation. R Vs L C Apply KVL: -V s +V R + V L +V C =0 Eq (1) Apply KCL: i R =i L = i C In Series Eq (2) Linear constitutive relations Resistor: V R =i R R Capacitor: V C =q C /C Inductor: dt di L V : note and Li L L L = = = & Substitute into Eq (1): -V s + i R R + dt di L L + q C /C =0 Note: we need to write the differential equation in terms of one variable, for example in terms of C q , charge in the capacitor. Relationship between charge and current: 2 dt q 2 d dt di therefore and i i i dt dq C L L R C C = = = = C q 2 dt q 2 d L dt dq R V- : Eq(1) in Thus C C C s = + + + s C C C V L 1 LC q dt q d L R 2 dt q 2 d = + + with L R n 2 and LC 1 2 n = = Eq (3) Recall mechanical system equation of motion equation of motion...
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This note was uploaded on 09/18/2011 for the course ME 140L taught by Professor Staff during the Fall '09 term at University of Texas at Austin.

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ME 140L - Lecture 3 - Second Order Systems - 1 MECHATRONICS...

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