timeresponse4 - + + (3) For T p = 5 s and the damping ratio...

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Time Response Q UESTION 4 For each pair of second–order system specifications below, find the pole locations. a. % OS = 12% and T s = 0.6 s b. % OS = 25% and T p = 5 s c. T s = 7 s and T p = 3 s For % OS = 12%, the damping ratio is 2 2 2 2 % 12 ln ln 100 100 0.5594 % 12 ln ln 100 100 OS OS ζ π - - = = = + + (1) For T s = 0.6 s and the damping ratio given in equation (1), the natural frequency is ( 29 4 rad 11.92 0.6 0.5594 s n ϖ = = (2) Therefore, the poles are located at 2 1 6.668 9.880 n n j j ζϖ - ± - = - ± . For % OS = 25%, the damping ratio is 2 2 2 2 % 25 ln ln 100 100 0.4037 % 25 ln ln 100 100 OS OS - - = =
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Unformatted text preview: + + (3) For T p = 5 s and the damping ratio given in equation (3), the natural frequency is 2 rad 0.6868 s 5 1 0.4037 n = =-(4) Therefore, the poles are located at 2 1 0.2773 0.6283 n n j j-±-= -± . Time Response For T s = 7 s, the product ζω n = 4/7 = 0.5714. For T p = 3 s 2 1 3 n p T π ϖ ζ-= = (5) Therefore, the poles are located at 2 1 0.5714 3 n n j j ζϖ-±-= -± . 2...
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This note was uploaded on 02/01/2012 for the course MECH ENG 279 taught by Professor Landers during the Spring '11 term at Missouri S&T.

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timeresponse4 - + + (3) For T p = 5 s and the damping ratio...

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