sol03 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Problem Set #3 Solution Posted: Friday, Sept. 28, ’07 1. A second–order system has the step response shown below. 1 Determine its trans- fer function. 0 0.5 1 1.5 2 t [sec] Answer: This is under–damped 2nd order system. Starting from the transfer function of the second order system ω 2 A n , s 2 +2 ζω n s + ω 2 we have to decide the parameters of A (constant), ζ (damping ratio) and ω n (natural frequency). From the final value theorem, 0 0.5 1 1.5 2 f(t) [a.u.] 1 2 lim s n = A s 0 s s 2 n s + ω n 2 1 a.u. denotes arbitrary units; its use appropriate when we consider a function that does not correspond to any particular physical quantity. 1
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± ² ³ ´ µ ·¸ and the steady state value is 1 (from the given figure). Therefore, A =1 . The step response of the under–damped second order system is 1 ae σ d t cos ( ω d t φ ) u ( t ) , where σ d = ζω n and ω d = ω n 1 ζ 2 . ζπ From the lecture note 7 (pp. 26), %OS = exp 1 ζ 2 : 72%. Thus the damping ratio ζ 0 . 1. To get the natural frequency, we choose two peak points at t 1 =0 . 35 sec and t 2 . 95 sec. The cosine term will be 1 at the peaks, so that we can consider exponential decay term only. f ( t 1 )=1 ae σ d t 1 . 72 f ( t 2 ae σ d t 2 . 4 Dividing the two equations, we obtain ae σ d t 1 1 1 . 72 = .
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.004 taught by Professor Derekrowell during the Fall '08 term at MIT.

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sol03 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

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