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Unit Impulse Respone

  1. Oct 12, 2013 #1
    Please see attached for problem and schematic

    I know from the question i should expect a decaying exponential funtion.

    Input: a unit impulse denoted in the laplace transform is 1
    outputs: -Kx0 - Cdx/dt

    -Kx0 - Cdx/dt +1 = 0

    factorise

    X0 ( -K - C*S) + 1 = 0

    -1 / ( -K - C*S ) = X0

    Is this correct up to now?
     

    Attached Files:

  2. jcsd
  3. Oct 12, 2013 #2
    No. You have been intermixing terms from the differential equation with terms from the LT equation. You need to start out by getting the differential equation correct. Express the differential equation in terms of x and δ(t). There shouldn't be an x0 in the differential equation, but there should be a δ(t).

    Chet
     
  4. Oct 13, 2013 #3
    So..

    δ(t) = -Kx - C(dx/dt) ?
     
  5. Oct 13, 2013 #4
    There should be plus signs on the RHS. Look back at your figure. After correcting this, what is the LT of the equation?

    chet
     
  6. Oct 13, 2013 #5
    im not sure but should it be

    δ(t) = -Kx + C(dx/dt)?

    My reasoning is that the spring will oppose the impulse and the damper will act in the same direction?

    If so, Taking laplace transforms ( Im very new to the topic )


    1 = -Kx + C*S*X0(S)?
     
  7. Oct 15, 2013 #6
    Vector force exerted by spring and damper on bar = [itex](kx+c\frac{dx}{dt})(-\vec{i_x})[/itex]
    Vector force by impulse on bar = [itex]δ(t)(+\vec{i_x})[/itex]
    Sum of forces on massless bar equal to zero:
    [tex](kx+c\frac{dx}{dt})(-\vec{i_x})+δ(t)(+\vec{i_x})=0\vec{i_x}[/tex]
    or
    [tex]-(kx+c\frac{dx}{dt})+δ(t)=0[/tex]


    If you're learning how to use Laplace transforms, then you have to learn it right. Is this a course in Laplace transforms? According to your book, what is the Laplace transform of the function dx/dt if x = x0 at t = 0?

    Chet
     
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