Solve ODE: Find Transfer Function G(S)

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math111
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Homework Statement


M[tex]\ddot{X}[/tex](t)+c[tex]\dot{X}[/tex](t)+kx(t) =f(t)
Initial Conditions:
x(0) = .02
[tex]\dot{X}[/tex](0)=0
-Use laplace transform to convert the ordinary differential equation in the time domain to an algebraic equation in the frequency domain.
-Derive the transfer Function G(S) = [tex]\frac{X(S)}{F(S)}[/tex]

Homework Equations


The Attempt at a Solution


mS[tex]^{2}[/tex]X(S) - .02MS + CSX(S) - .02C + KX(S) = F(S)

[mS[tex]^{2}[/tex] - CS+K]X(S) = F(S) +.02MS - .02C

X(S) = F(S) +.02MS - .02C / mS[tex]^{2}[/tex] - CS+K

This is where I get confused.
1. Should I of divided out the M in the beginning?(i.e. k/m, c/m..)
2. At this point do I need partial fractions to go further?
 
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math111 said:
X(S) = F(S) +.02MS - .02C / mS[tex]^{2}[/tex] - CS+K
Recheck this line. You should not have any minus sign.

This is where I get confused.
1. Should I of divided out the M in the beginning?(i.e. k/m, c/m..)
2. At this point do I need partial fractions to go further?
A pity you aren't given the unknowns explicitly. Because using the quadratic formula to get the factors looks really complicated. I really don't see how to use partial fractions since you're not given F(s).
 
Defennder said:
Recheck this line. You should not have any minus sign.

A pity you aren't given the unknowns explicitly. Because using the quadratic formula to get the factors looks really complicated. I really don't see how to use partial fractions since you're not given F(s).

yeah at 3am I might make mistakes...
it should be
X(S) = [F(S) +.02MS + .02C] / [MS^2 + CS+K]

Now I see X(S) = [F(S)/M]/[S^2 + CS/M+K/M] + [.02S + .02C/M]/[S^2 + CS/M+K/M]

from here I need some more help.. I think C/M and K/M mean something else..