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2nd order Laplace transforms

  1. Jul 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve the DE for y(t) with the IC's
    y(0)=20.8m/s and y'(0)=0

    if the input is a step function scaled by the desired velocity Vo.
    vd(t)=Vou(t).
    Assume the desired velocity Vo=27.8m/s



    2. Relevant equations

    y''(t) + (D/M)y'(t) + (K/M)y(t) = (K/M)vd(t)

    M = 1,000kg
    D = 100kg/s
    K = controller gain
    y(t) = output velocity
    vd is the input function



    3. The attempt at a solution


    So I'm Laplace transforming the whole 2nd order equation and I end up with a mess. The next problem is to find the optimal controller gain K for a desired response.

    My Laplace transform of the 2nd order equation is:

    Y(s) = [ KVo + Ms2y(0) + Dsy(0) ] / [ s2 + (D/M)s + (K/M) ]

    if in fact i'm doing it correctly, here is where i am stuck.
     
    Last edited: Jul 20, 2013
  2. jcsd
  3. Jul 20, 2013 #2
    plugging in the variables into Y(s) i get:


    Y(s) = [ 27.8K + 20800s2 + 2080s ] / [ s2 + .1s + .001K ]
     
  4. Jul 20, 2013 #3
    This Laplace Transform doesn't look correct to me. Please show your work. Please show your Laplace Transform expressions for y'', y', and vd(t).

    Chet
     
  5. Jul 20, 2013 #4
    Laplace Transform expressions for y'', y', and vd(t).


    y''(t):

    s2Y(s) - sy(0) - y'(0)

    +

    (D/M)y'(t):

    (D/M)(sY(s) - y(0))

    +

    (K/M)y(t):

    (K/M)Y(s)

    =

    (K/M)vd(t):

    (K/M)(Vo/s)


    then I factor out Y(s) and then solve for Y(s). that's when i get that ugly fraction up there that i don't know what to do with. i know how to do simple inverse transforms with tables and partial fractions.
     
    Last edited: Jul 20, 2013
  6. Jul 20, 2013 #5
    It looks like the expressions above are correct, but there were algebra errors in solving for Y(s). Please try again. It looks like you are missing a factor of Ms in the denominator.
     
  7. Jul 20, 2013 #6
    ok so this time i got:

    Y(s) = [ (KVo/Ms) + y'(0) + sy(0) + (D/M)y(0) ] / [ s2 + (D/M)s + (K/M) ]
     
  8. Jul 20, 2013 #7
    Good. Don't forget that y'(0) = 0.
    Now, the next step is to manipulate this into a form that is a linear combination of some of the transforms in your tables. Start out by factoring y(0)/s out of the numerator.

    Chet
     
  9. Jul 20, 2013 #8
    if it wasn't for that darn K. so:


    (y(0)/s)[ (KVo/My(0)) + s2 + (D/M)s ] / [ s2 + (D/M)s + (K/M) ]
     
  10. Jul 21, 2013 #9
    No problem. I'm going to retype what you have:

    [tex]y(s)=\frac{y(0)}{s}\frac{(s^2+(D/M)s+\frac{KV_0}{My(0)})}{(s^2+(D/M)s+\frac{K}{M})}[/tex]

    Check out the terms in parenthesis in the numerator and denominator. Does this suggest something you can do algebraically to simplify things?
     
  11. Jul 21, 2013 #10
    the only option i can see would be to pull 1 s out of the first 2 terms of each polynomial. if the K weren't there i could factor the polynomial using quadratic.
     
  12. Jul 22, 2013 #11
    Suppose you wrote the term in parenthesis in the numerator as
    [tex]\left(s^2+(D/M)s+\frac{KV_0}{My(0)}\right)=\left(s^2+(D/M)s+\frac{K}{M}\right)+\left(\frac{KV_0}{My(0)}-\frac{K}{M}\right)[/tex]

    Then the Laplace Transform would become:

    [tex]y(s)=\frac{y(0)}{s}\left(1+\frac{\left(\frac{KV_0}{My(0)}-\frac{K}{M}\right)}{(s^2+(D/M)s+\frac{K}{M})}\right)[/tex]

    This should be pretty easy to invert.

    Chet
     
  13. Jul 29, 2013 #12
    A big big big thank you to you guys for your help. i ended up with an A on the project and a B for the semester. Thank you!
     
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