Chap3StudentSolutions

15 is then the convolution z t 015 u t h t or z

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Unformatted text preview: pension model Using the basic principle, F = ma , we can write Solutions 3-39 M. J. Roberts - 8/16/04 K s[ y( t) − x( t) − d0 ] + K d or (a) d [y(t) − x(t)] + mg = − m y′′(t) dt m y′′ ( t) + K d y′ ( t) + K s y( t) = K d x′ ( t) + K s x( t) + K sd0 − mg . At rest all the derivatives are zero and K s ( y( t) − x( t) − d0 ) + mg = 0 . Solving, y( t) − x( t) = (b) K sd0 − mg 75000 × 0.6 − 1500 × 9.8 = = 0.404 m Ks 75000 The describing equation is m y′′ ( t) + K d y′ ( t) + K s y( t) = K d x′ ( t) + K s x( t) + K sd0 − mg . which can be rewritten as or m y′′ ( t) + K d [ y′ ( t) − x′ ( t)] + K s[ y( t) − x( t)] − K sd0 + mg = 0 mg =0 m y′′ ( t) + K d [ y′ ( t) − x′ ( t)] + K s y( t) − x( t) − d0 + Ks Let z( t) = y( t) − x( t) − d0 + or mg . Then y′′ ( t) = z′′ ( t) + x′′ ( t) and Ks m[z′′ ( t) + x′′ ( t)] + K d z′ ( t) + K s z( t) = 0 m z′′ ( t) + K d z′ ( t) + K s z( t) = − m x′′ ( t) This equation is in a form which describes an LTI system. We can find its impulse response. After time, t = 0, the impulse response is the homogenous solution. The eigenvalues are 2 −K d ± K d − 4 mK s K λ1,2 = =− d ± 2m 2m 2 Kd Ks = −6.667 ± j 2.357 . 2− m 4m The homogeneous solution is h( t) = K h1eλ 1 t + K h 2eλ 2 t = K h1e( −6.667 + j 2.357) t + K h 2e( −6.667 − j 2.357) t . Since the system is underdamped another (equivalent) form of homogeneous solution will be more convenient, h( t) = e −6.667 t [K h1 cos(2.357 t) + K h 2 sin(2.357 t)] . Solutions 3-40 M. J. Roberts - 8/16/04 The impulse response can have a discontinuity at t = 0 and an impulse but no higher-order singularity there. Therefore the general form of the impulse response is h( t) = Kδ ( t) + e −6.667 t [K h1 cos(2.357 t) + K h 2 sin(2.357 t)] u( t) Integrating both sides of the describing equation between 0 − and 0 + , ( m h′ (0 ) − h′ (0 + 0+ − )) + K (h(0 ) − h(0 )) + K ∫ h(t)dt = 0 . + − d s 0− (The integral of the doublet, which is the derivative of the impulse excitation, is zero.) Since the impulse response and all its derivatives are zero before time, t = 0, it follows then that 0+ m h′ (0 ) + K d h(0 ) + K s ∫ h( t) dt = 0 + + 0− and m(−6.667K h1 + 2.357K h 2 ) + K d K h1 + K sK = 0 . Integrating the describing equation a second time between 0 − and 0 + , 0+ m h(0 ) + K d ∫ h( t) dt = 0 + 0− or mK h1 + K d K = 0 . Integrating the describing equation a third time, 0+ m ∫ h( t) dt = − m 0− or mK = − m ⇒ K = −1 . Solving for the other two constants, K h1 = Kd and m K K m −6.667 d + 2.357K h 2 + K d d − K s = 0 m m or 2 Ks Kd K − 2 + 6.667 d m Kh 2 = m m 2.357 Kd m Therefore h( t) = −δ ( t) + e −6.667 t [13.333 cos(2.357 t) − 16.497 sin(2.357 t)] u( t) Solutions 3-41 M. J. Roberts - 8/16/04 (c) The response to a step of size 0.15 is then the convolution, z( t) = 0.15 u( t) ∗ h( t) or ∞ { } z( t) = 0.15 ∫ −δ...
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