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Laplace Transform of unit step function HELP

  1. Feb 22, 2010 #1
    Laplace Transform of unit step function HELP!!!

    1. The problem statement, all variables and given/known data

    f(t)= e^t on 0<=t<1
    . . . . t on 1<=t<2
    . . . . sin(t) on 2<=t<infinity

    2. Relevant equations

    Unit Step Function

    3. The attempt at a solution

    Here is my attempt at a solution...

    f(t)=e^t*u(t)-e^t*u(t-1)+t*u(t-1)-t*u(t-2)+sin(t)*(t-2)

    Now I know I need to transform this equation, but I'm not sure how to break it up into something a little easier to comprehend. Any help would be greatly appreciated!!! Thanks in advance!
     
  2. jcsd
  3. Feb 23, 2010 #2

    rock.freak667

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    Re: Laplace Transform of unit step function HELP!!!

    Your new expression for f(t) is correct.


    But you should know that L[f(x-a)H(x-a)]=e-asF(s) where F(s)=L[f(t)]
     
  4. Feb 23, 2010 #3
    Re: Laplace Transform of unit step function HELP!!!

    So do I just do...

    L[e^t*u(t)]-L[e^t*u(t-1)]+L[t*u(t-1)]-L[t*u(t-2)]+L[sin(t)*(t-2)] ??

    I don't feel like that is what I'm supposed to do. I think I need to bring some things out and put them in front of the e^t, t, and sin(t) but I have a hard time knowing what to bring out and then exactly how to transform that.
     
  5. Feb 23, 2010 #4
    Re: Laplace Transform of unit step function HELP!!!

    You need to evaluate each laplace transform now.
    For example the Laplace transform of sin(t)
    L[sin(omega*t)] = omega / ( s^2 + omega^2)
    You also have to use the properties of laplace transforms especially the time shift property.

    Or you can just do the integrals for each, you get the same answer and it's about the same amount of work.
     
    Last edited: Feb 23, 2010
  6. Feb 23, 2010 #5

    vela

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    Re: Laplace Transform of unit step function HELP!!!

    This is the first step in one possible approach to the problem. Now you need to evaluate what each term is, and that's where you do what you're talking about. For example, let's consider the second term, [itex]L[e^t u(t-1)][/itex]. As rock.freak noted, you have the property [itex]L[f(t-a)u(t-a)]=e^{-as}F(s)[/itex] satisfied by the Laplace transformation. The second term is almost of the right form but not quite because it contains [itex]f(t)=e^t[/itex] rather than [itex]f(t-1)=e^{t-1}[/itex]. You can however use the fact [itex]e^t = e^{(t-1)+1}=e^{t-1} e^1[/itex] to get

    [tex]L[e^t u(t-1)] = L[e e^{t-1}u(t-1)]=e L[e^{t-1}u(t-1)] = e\frac{e^{-s}}{s-1} = \frac{e^{-s+1}}{s-1}[/tex]

    You can use similar tricks to evaluate the rest of the terms.
     
  7. Feb 23, 2010 #6
    Re: Laplace Transform of unit step function HELP!!!

    Thank you very much! This helps tremendously!!!
     
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