1. Not finding help here? Sign up for a free 30min 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!

Laplace transformation help

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data[/b]
    NruJm.png
    I'm required to find the current i0(t), i wrote KVL equations for each loop, I0 can be expressed interms of i1 and i3, i3 being the current source on the right. i managed to get the solution down to the following :

    i0 = (s + 4)/(3*s^2 + 4*s + 1)

    I simplified the denominator by completing the square and got:

    i0 = (s + 4)/((3*(s+4/6)^2 - 12/36))

    Now i'm kind of stuck, i one of solutions is in the form (s+a)/((s+a)^2 + w^2) I'm not sure how i can get my solution to be similiar to this, any tips would be awesome :)
    thankyou!



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 21, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Can you show your initial loop equations? I'm not seeing the same result that you have for io.

    (Try using the x2 and x2 icons in the edit panel header to produce exponents and subscripts :wink: )
     
  4. May 21, 2012 #3
    hmm i think i made a few mistakes, i tried it again and this is what i get :

    kvl 1:
    -4/s + i1 + i1/s - i0/s
    => 4/s = i1 + i1/s - i0/s

    kvl 2:
    2io + io - 1/(s+1) + i0/s - i1/s = 0

    Now i rearranged kvl 2 to find i1 and i got:
    i1/s = 3i0 - 1/(s+1) + i0/s
    = i0(3 + 1/s) - 1/(s+1)
    i1 = s(i0(3+1/s) - 1(s+1)) = i0(3s + 1) - s/(s+1)

    so how i got my original answer is a bit of a mystery haha. I tried it again and substituted i1 into kvl 1 and got the answer from wolphram alpha

    (s+4)/(3s2 + 4s)

    Is this what you got?
     
  5. May 21, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Yup, that's what I found.
     
  6. May 21, 2012 #5
    Ok, now that i got that right :P i need to convert it back to the time domain, but no solution in my list seems to match. any ideas?
     
  7. May 21, 2012 #6
    ahh, the answer was so simple, i used the inverse laplace transformation

    the equation is in the form:

    s+4 = A/s + B/(3s+4)
    A = 1
    B = -2

    1/s = u(t)
    B/(3(s+4/3) = -2e-(4/3)t*(3)

    i'm not to sure do i just multiply the 3 to the equation?
     
  8. May 21, 2012 #7

    gneill

    User Avatar

    Staff: Mentor

    I suspect that the 3 should divide the equation, not multiply. And the u(t) should be transformed, too.
     
  9. May 21, 2012 #8
    well A = 1, 1/s = u(t) so what else could i do to that?
    my final solution was:

    1/u(t) - (2/3)*e-(4/3)t/(s+(4/3))
     
  10. May 21, 2012 #9

    gneill

    User Avatar

    Staff: Mentor

    There should be no s's in the time domain result, only t's and constants. If time is assumed to begin at t=0 for the circuit, the u(t) becomes a constant 1.
     
  11. May 21, 2012 #10
    Oh ok i think i follow, i realised i had a small mistake (not sure why i had the extra s ahah) it should be:

    u(t) - (2/3)*e-(4/3)t

    i need to find I for t>0, so at t>0 u(t) = 1, is that what you're implying?

    so my answer should be:

    1 - (2/3)*e-(4/3)t
     
  12. May 21, 2012 #11

    gneill

    User Avatar

    Staff: Mentor

    That looks good to me :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace transformation help
  1. Laplace Transform help (Replies: 1)

Loading...