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: Help with an easy Laplace Transform problem.

  1. Sep 30, 2008 #1

    If, f(t) = d/dt [(e^-5t) (cos2t)]

    Find F(s).

    Eh, well, I don't really know what to do, can I get some pointers?

    Am I supposed to integrate both sides, so I can get rid of d/dt, and then apply the integration property to find the answer?
    Last edited: Sep 30, 2008
  2. jcsd
  3. Sep 30, 2008 #2


    User Avatar
    Homework Helper

    You should

    • Know the transform of [tex] \cos 2t [/tex]
    • Have a theorem that tells you how to find the transform of a derivative of a function
    • Know how multiplication by an exponential function influences the transform of a given function

    Using these ideas, not necessarily in the order presented here, will get you through the problem.
  4. Sep 30, 2008 #3
    Ok, thank you statdad.

    Anyway, first of all I made a typo on the initial equation. The correct f(t) is;

    f(t) = d/dt [(e^-5t) (cos2t)]

    Now, what I know is;

    if f(t) = cos2t, then F(s) = s / (s^2 + 4)

    also, if f(t) = (e^-5t) (cos2t) ,then F(s) = (s + 5) / [(s + 5)^2 +4]

    Now, I know that;

    df(t) / dt = sF(s) - f(0) and that f(0) = 1

    Now, I will try to add everything together and find an answer..
  5. Sep 30, 2008 #4
    Umm, well;

    I know that df(t) / dt = sF(s) - f(0) and that f(0) = 1


    L[f(t)] = s [(s + 5) / (s + 5)^2 +4] - 1

    F(s) = [s(s+5) / (s+5)^2 + 4 ] - s

    Am I correct?
    Last edited: Sep 30, 2008
  6. Oct 1, 2008 #5


    User Avatar
    Homework Helper

    Where did this 's' come from? More importantly does the question say that f(0) = 1?
  7. Oct 1, 2008 #6
    You are correct, I can not say that f(0) = 1, also, that was another mistake(the s). But, well, I guess, that doesn't help me much. :x

    Anyway, I found a ti89 program to find the transform. I know now that the answer is;

    F(s) = -(5s + 29) / s^2 + 10s + 29

    But, huh, no idea how to get there. It's 2:35 am, being doing all kind of problems with laplace transforms, and the first problem(this one), I cant solve it. Woot, for some reason I feel this problem or something really similar to this is coming on my exam tomorrow. >_>
  8. Oct 1, 2008 #7
    So, umm f(0) should be zero, right?

    f(0) = d/dt ( cos0 ) = 0

    Ok, now, I get;

    F(s) = [s(s+5) / (s+5)^2 + 4 ]

    F(s) = s(s+5) / s^2 + 10s + 29

    Hey, well at least the denominator is correct. >.> Wonder what am I doing wrong.
  9. Oct 1, 2008 #8


    User Avatar
    Homework Helper

    You're right about that. f(0) = 1 by substitution. I thought we were solving a DE and the initial condition was omitted.

    Your answer is correct. Just combine the -1 into the same fraction and you'll get what your program gave you.
  10. Oct 1, 2008 #9
    Alright. Thank you, Defender. Now I can sleep in peace.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook