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An inverse Laplace transform

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    .......find the inverse transform of the given function.

    F(s)= (2s+2)/(s2+4s+5)

    2. Relevant equations

    On another page, I see these 2:


    3. The attempt at a solution

    This is my first time doing anything with Laplace transforms.

    I can change the denominator to (s+2)2 + 1 and factor a -2 out of the numerator. Then it looks like the relevant equations.


    But I don't understand where to go from here. The -2, of course, is just a constant that can chill out front. Judging from the answer in the back of the book, I use a combination of both relevant equations. Explain the algebra involved here.
  2. jcsd
  3. Mar 9, 2010 #2


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    Ok, so a=(-2) and b=1, right? The numerator of the first expression is 1 and the second is s+2. Looks like you want to solve c*1+d*(s+2)=(2s+2) for c and d. Then multiply your first f(t) by c and the second f(t) by d.
  4. Mar 9, 2010 #3
    Let's slow down for a second.

    I have (s-1/2)/((s+2)2+1).

    That isn't in the form b/((s-a)2+b2) because the b's are inconsistent. If I make a=-2, and b=1, the denominator works out, but then the numerator doesn't.

    It also isn't in the form (s-a)/((s-a)2+b2) because the a's are inconsistent. If I do the same, the numerator doesn't work out.

    Explain this to me as if I was a 5-year-old.

    http://blog.oregonlive.com/petoftheday/2008/04/large_thanks.bmp [Broken]
    Last edited by a moderator: May 4, 2017
  5. Mar 10, 2010 #4


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    Ok, five year old. You got the a and b figured out. Multiply each of your t expressions by a separate constant. So the laplace transform of c*e^(-2t)*sin(t) is c*1/((s+2)^2+1) and the laplace transform of d*e^(-2t)*cos(t) is d*(s+2)/((s+2)^2+1). If you add those you get that the laplace transform of e^(-2t)*(c*sin(t)+d*cos(t)) is (c*1+d*(s+2))/((s+2)^1+1). Now you just have to figure out c and d so that (c*1+d*(s+2))=2*s+2.
  6. Mar 10, 2010 #5
    Got it!

    Thanks, bro!
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