# Note there are two unknowns x1 t and x 2 t since we

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Unformatted text preview: + k 2 x 2 = k 2 x1 (2) . Note there are two unknowns, x1 (t ) and x 2 (t ) . Since we are interested in x 2 (t ) , we eliminate x1 (t ) by substituting it from Equation (2) into Equation (1). Taking LT, assuming alli.c.’s are zero we have [m s 1 2 + b s + k ]X + b1 s + (k1 + k 2 ) X 1 ( s ) = F ( s ) + k 2 X 2 ( s ) [m s 2 2 2 2 2 ( s) = k 2 X 1 (s ) (1a) (2a) Substituting X1(s) from Equation (2a) into Equation (1a), m2 s 2 + b2 s + k 2 X 2 ( s ) 2 m1 s + b1 s + (k1 + k 2 ) = F (s ) + k 2 X 2 (s) k2 [ [m s 2 1 {[m s 1 2 [ [ [ + b1 s + (k1 + k 2 ) m2 s 2 + b2 s + k 2 X 2 ( s ) − k 2 X 2 (s ) = F ( s) k2 2 + b1 s + (k1 + k 2 ) m2 s 2 + b2 s + k 2 − k 2 }X 2 ( s ) = k 2 F ( s ) X 2 ( s) k2 = 2 2 F ( s) m1 s + b1 s + (k1 + k 2 ) m2 s 2 + b2 s + k 2 − k 2 [ [ X 2 ( s) k2 = 4 3 F (s) m1m2 s + (m1b2 + m2 b1 ) s + [m1 k 2 + m2 (k1 + k 2 )]s 2 + [b1 k 2 + b2 (k1 + k 2 )]s + k1 k 2 Notes on common mistakes: (1) All ic’s = 0 in deriving TF. (2) TF should not contain any input or output variables. [2]...
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## This note was uploaded on 10/15/2012 for the course ME 3015 taught by Professor Ueda during the Spring '08 term at Georgia Institute of Technology.

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