timeresponse3 - F s = 1 s and X s is 29 29 29 29 29 2 2 2 2...

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Time Response Q UESTION 3 Solve for x ( t ) in the system in Figure 3.1 if f ( t ) is a unit step. Assume zero initial conditions. 1 kg 1 N/(m/s) 5 N/m x(t) f(t) Figure 3.1 Applying Newton’s second law to the mass ( 29 ( 29 ( 29 ( 29 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 X s F s s s = + + (2) If the applied force is a unit step,
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Unformatted text preview: F ( s ) = 1/ s and X ( s ) is ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 1 0.2 1 0.2 5 5 0.2 0.5 0.2 4.75 0.5 0.2 4.75 0.5 4.75 0.5 4.75 s X s s s s s s s s s s s + = =-+ + + + + =--+ + + + (3) Taking the inverse Laplace Transform of equation (3) ( 29 ( 29 ( 29 ( 29 0.5 0.5 0.2 0.5 0.2 sin 4.75 0.2 cos 4.75 4.75 t t x t e t e t--=--(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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