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Solving driven differential equations

  1. Sep 20, 2010 #1
    1. The problem statement, all variables and given/known data

    The suspension system of a car is designed so that it is a damped system described by

    z'' + 2z' + z = f(t)

    where z is the vertical displacement of the car from its rest position. The car is driven over a (smooth!) road which has "catseye" embedded in the road surface every d metres. Assume that each time the car passes over a catseye, the suspension system receives a unit impulse. Thus, if the car is driven with velocity v m/s over K + 1 catseyes then

    f(t) = (see attachment)


    (a) Find z(t) when z(0) = z'(0) = 0.


    (b) Plot the solutions for K = 10 and K = 50 with v = 10 What is the eff ect of a larger K? Fix K = 50. Let v = 10 and plot the solutions for d = 5 and d = 10 on the same axes. What is the e ffect of the distance between the cateyes? Let d = 10 and plot the solutions for v = 10 and v = 20. What is the e ffect of speed?



    2. Relevant equations
    on attachment
     

    Attached Files:

  2. jcsd
  3. Sep 20, 2010 #2

    gabbagabbahey

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    What have you tried?
     
  4. Sep 20, 2010 #3
    Ive got no idea where to start. Thats the problem. I would assume f(t) needs to be expressed in terms of heaviside step functions and then the differential equation solved but im not sure.
     
  5. Sep 20, 2010 #4

    gabbagabbahey

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    Have you learned the method of using Laplace transforms to solve ODEs? If so, try that...
     
  6. Sep 20, 2010 #5
    Ok thats all well and good but can you explain a little bit further. We have covered laplace transforms but only recently and I am trying to work thorugh some problems to understand how it works. Thanks
     
  7. Sep 20, 2010 #6

    gabbagabbahey

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    The first step is always to take rthe Laplace transform of both sides of the ODE....what do you get when you do that?
     
  8. Sep 20, 2010 #7
    The left side is easy if you set it to equal zero. The laplace transform is

    s^2*Y(s) + 2s*Y(s) + Y(s) = 0

    but im not sure about the right hand side with the sum from 0 to K of the dirac delta function.
     
  9. Sep 20, 2010 #8

    gabbagabbahey

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    Do you know how to take the Laplace transform of a single delta function? If so, just use the fact that the Laplace transform is linear:

    [tex]\mathcal{L}\left[g_1(t)+g_2(t)\right]=\mathcal{L}\left[g_1(t)\right]+\mathcal{L}\left[g_2(t)\right][/tex]
     
  10. Sep 20, 2010 #9
    Im not sure about that part. Can you elaborate.
     
  11. Sep 20, 2010 #10

    gabbagabbahey

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    You're not sure how to take the LT odf a delta function? Or you aren't sure how to add a bunch together?
     
  12. Sep 20, 2010 #11
    Im not sure how to take the laplace transform of δ(t-kd/v)
     
  13. Sep 20, 2010 #12

    gabbagabbahey

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    It's probably in your textbook, so you might try opening that up and reading part of it :wink:....or, you can use the definition of Laplace transform and the definition of delta function and calculate it directly....
     
  14. Sep 21, 2010 #13
    Alright after some time I get the laplace transform of the Right hand side as e-(kd/v*s)

    and then sloving the entire equation for Y(s) I get

    Y(s) = e-(kd/v*s)/s2+2s+1

    But does this take into account the sumation and then how do I take the inverse laplace transform to solve the DE
     
  15. Sep 21, 2010 #14

    vela

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    No, it doesn't take into account the summation. The Laplace transform of the righthand side is

    [tex]L[f(t)] = L\left[\sum_{k=0}^K \delta(t-kd/v)\right] = \sum_{k=0}^K L[\delta(t-kd/v)][/tex]

    where the last step uses the linearity of the Laplace transform. What you calculated is the inside of the last summation.

    The exponential factor merely indicates a delay. To find the inverse Laplace transform, you temporarily neglect it, find the inverse transform of the rest, and then time-shift the answer. For example, suppose X(s)=e2s/(s+1). To find x(t), you first find the inverse Laplace transform of 1/(s+1), which is e-t. Now you time-shift the answer by 2, so x(t)=e-(t-2)u(t-2).
     
  16. Sep 21, 2010 #15
    I find

    z(t) = t e-(t-kd/v) Heaviside(t-kd/v)

    The only thing im not sure about is fitting the summation back in. How do I do this??
     
  17. Sep 22, 2010 #16

    gabbagabbahey

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    Close, but shouldn't your [itex]t[/itex] out front really be [itex]\left(t-\frac{kd}{v}\right)[/itex]?

    Not only is the Laplace transform linear, so is its inverse. That means if [tex]Z(s)=\sum_{k=0}^K g_k(s)[/itex] , then

    [tex]z(t)=\mathcal{L}^{-1}\left[\sum_{k=0}^K g_k(s)\right]=\sum_{k=0}^K \mathcal{L}^{-1}\left[g_k(s)\right][/tex]
     
    Last edited: Sep 22, 2010
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