solution08 - 1/. period = 12.66 / 4 = 3.16 secs static gain...

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1/. sec / rads 985 . 1 / 2 secs 16 . 3 4 / 66 . 12 = = = = period period d π ω static gain = Input Ouputs k / 20 / 200 = logarithmic decrement = 6413 . 0 11 .. 1 44 . 144 ln 4 1 ln 1 1 = = = n X X n δ 981 . 3 4052 . 0 81 . 39 2 ) ( 9953 . 1 1 / 101539 . 0 4 2 2 2 2 2 2 2 + + = + + = = = = + = s s s s k s H n n n d n ζω ζ This is pretty close. The graph was created using 4 4 . 0 40 ) ( 2 + + = s s s H
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2/. Differential equation for system: x m k x m c y m k y m c y + = + + Consider instead the “standard” system: z m y m k y m c y 1 = + + where kx x c z + = For the values given, the parameters of this system are: m/N 001 . 0 gain static 25 . 1 10 100 25 1 . 0 ) ( 2 = = = = + + = k s s s H n ζ ω When x ( t ) is 0.02 u ( t ) metres, z ( t ) is the sum of an impulse (note that the derivative of a step is an impulse) and a step. ) ( Newton 20 ) ( seconds - Newton 5 ) ( ) m 02 . 0 )( N/m 1000 ( ) ( ) m 02 . 0 )( Ns/m 250 ( t u t t u t kx x c z + = + = + = δ The response of an overdamped second order system to an impulse (p5 -11) is: [] t t N N N e e C k t y ) 1 ( ) 1 ( 2 2 2 1 2 ) ( + = Plugging in C = 5 N-sec and the other values gives: t t e e t y 20 5 3 033 . 0 3 033 . 0 ) ( = The response of an overdamped second order system to a step (p5-13) is:
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This note was uploaded on 07/16/2009 for the course SYSC 3600 taught by Professor John bryant during the Winter '08 term at Carleton CA.

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solution08 - 1/. period = 12.66 / 4 = 3.16 secs static gain...

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