This preview shows page 1. Sign up to view the full content.
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]...
View
Full
Document
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.
 Spring '08
 UEDA
 Mechanical Engineering

Click to edit the document details