Timeresponse11 - and 2.5 0.5 5 3.618-= If the applied force is a unit step F s = 1 s and X s is 29 29 2 1 0.2 0.3236 0.1236 1.382 3.618 1.382 3.618

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Time Response Q UESTION 11 Solve for x ( t ) in the system in Figure 11.1 if f ( t ) is a unit step. Assume zero initial conditions. 1 kg 5 N/(m/s) 5 N/m x(t) f(t) Figure 11.1 Applying Newton’s second law to the mass ( 29 ( 29 ( 29 ( 29 5 5 x t x t x t f t = - - + (1) Taking the Laplace Transform, assuming zero initial conditions, the transfer function is ( 29 ( 29 2 1 5 5 X s F s s s = + + (2) The poles are 2.5 0.5 5 1.382 - + = -
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Unformatted text preview: and 2.5 0.5 5 3.618--= -. If the applied force is a unit step, F ( s ) = 1/ s and X ( s ) is ( 29 ( 29 2 1 0.2 0.3236 0.1236 1.382 3.618 1.382 3.618 5 5 A B C X s s s s s s s s s s = = + + =-+ + + + + + + (3) Taking the inverse Laplace Transform of equation (3) ( 29 1.382 3.618 0.2 0.3236 0.1236 t t x t e e--=-+ (4)...
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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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