1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
    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).

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


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



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







    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.

  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:


    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

    Then the Laplace Transform would become:


    This should be pretty easy to invert.

  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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted