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Help with an easy Laplace Transform problem.

  1. Sep 30, 2008 #1
    Problem;

    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

    statdad

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    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

    so,

    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

    Defennder

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    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

    Defennder

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    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.
     
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