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### pset3_sol

Course: ME 3015, Fall 2011
School: Georgia Tech
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Word Count: 675

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PROBLEM ME3015 SET #3 (25 points). Solutions 1. (6 points) Assume that the system is initially in equilibrium. The gravitational force of mass causes a static deflection of the spring. The external force u applies as input and the displacement y is the output of the system. Find the state-space representation of the spring-mass-pulley shown below. Then convert the state-space representation to the corresponding...

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PROBLEM ME3015 SET #3 (25 points). Solutions 1. (6 points) Assume that the system is initially in equilibrium. The gravitational force of mass causes a static deflection of the spring. The external force u applies as input and the displacement y is the output of the system. Find the state-space representation of the spring-mass-pulley shown below. Then convert the state-space representation to the corresponding transfer function from u to y. Moment of Inertia J R Neutral length m k without MASS u mg (Origin: Equilibrium position) y Solution Define the tension acting in the wire as T=T + k where k=mg. Since the displacement y is defined from the equilibrium position, the effect of gravity is not included in the dynamic equations. T m = && y m u & J& TR = kR 2 The equations of motion for the mass and the pulley are given as m&& = T + u y & J& = TR kR 2 Since y = R , the equation of motion for the pulley can be written as J&& = TR 2 kR 2 y y By eliminating T from the equations, we have J&& = (u m&&) R 2 kR 2 y . y y Thus, && + y kR 2 R2 u y= J + mR 2 J + mR 2 Define state variables as x1 = y & x2 = y The output y is given by y = x1 . A state-space representation is given by 0 & x1 kR 2 = x &2 J + mR 2 x y = [1 0] 1 x2 1 x 0 1 + R 2 u 0 x2 J + mR 2 ANS Transfer function G(s) = Y (s) R2 = (ANS) U ( s) ( + J mR 2 ) s 2 + kR 2 2. (6 points) 1 0 1 0 G ( s ) = C ( sI A) B = [1 0] sI a b 1 1 1 s 1 0 = [1 0] a s b 1 s b 1 0 1 [1 0] = a s 1 s ( s b) a 1 1 [1 0] s s bs a 1 =2 s bs a = 2 ANS 3. (6 points) Obtain the mathematical model (differential equation) of the circuit shown below: L R3 R1 C R2 e(t) i1 From the circuit diagram we obtain di1 + R1i1 + R2 (i1 i2 ) = e dt ANS 1 R3i2 + i2 dt + R2 (i2 i1 ) = 0 C2 L i2 4. (7 points) Find the transfer function from Ei(s) to EO(s) for the electrical system shown below: C R1 R3 ei(t) R4 R2 eo(t) Solution i4 i1 R1 i3 A i2 ei(t) C i5 R3 R2 R4 eo(t) e A : voltage at A from the GND. i1 = ei e A R1 i2 = eA R2 i3 = e A eO R3 i4 = C i5 = d (e A eO ) dt eO R4 Since i1 = i2 + i5 we have ei e A e A eO . = + R1 R2 R4 Also since i4 + i3 = i5 we have (1) C d e e e (e A eO ) + A O = O R3 R4 dt (2) From (1), ei 1 e 1 = + e A + O R R R1 1 R4 2 Taking Laplace transformation of this equation, we obtain Ei 1 EO 1 = + R R EA + R R1 1 2 4 (3) Also taking Laplace transformation of (2) we obtain Cs ( E A EO ) + E A EO EO = R3 R4 From the above, EA = R3 R4Cs + R3 + R4 EO R4 ( R3Cs + 1) By substituting (4) into (3), Ei R1 + R2 R3 R4Cs + R3 + R4 EO = R R R ( R Cs + 1) EO + R R1 1 2 4 3 4 Therefore, EO R2 R4 ( R3Cs + 1) ANS = Ei [R1 R2 R3 + ( R1 + R2 ) R3 R4 ]Cs + R1 R2 + ( R1 + R2 )( R3 + R4 )
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