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Laplace tranforms

  1. May 7, 2009 #1

    i.e L{u(t)} = 1/s
  2. jcsd
  3. May 7, 2009 #2


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    The unit step function is defined as::

    u(t)=\begin{cases} 0, & t < 0 \\ 1, & t \ge 0 \end{cases}

    Now take the Laplace transform.

    L[u(t)]=\int_0^\infty u(t) e^{-st} dt=\int_0^\infty 1*e^{-st} dt

    Because on the interval [itex]0 \leq x < \infty, u(t)=1[/itex].

    You should be able to work it out now.

    The same holds for the two-sided Laplace transform, because on the interval [itex]-\infty<x<0[/itex] the unit step function is 0.
  4. May 7, 2009 #3


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    Perhaps what you really want is L(u(t-a)).

    [tex]u(t-a)=\begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases}[/tex]

    [tex]L(u(t-a))= \int_0^\infty u(t-a)e^{-st} dt= \int_a^\infty e^{-st}dt[/tex]
    Let v= t- a. Then t= v+ a, dv= dt, when t= a, v= 0, and when t= [itex]\infty[/itex], v= [itex]\infty[/itex]. The integral becomes
    [tex]L(u(t-a)= \int_0^\infty e^{s(v+a)}dv= \int_0^\infty e^{sv}e^{sa}dv[/tex]
    [tex]= e^{sa}\int_0^\infty e^{sv}dv= e^{sa}e^{-st}dt[/tex]
    and that last integral is the Laplace transform of 1.

    That u(t-a) step function multiplying a function basically multiplies the Laplace transform of the function by [itex]e^{as}[/itex].
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