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Inverse laplace of 1s

  1. Feb 5, 2013 #1
    This is very tricky for me. How to find the inverse laplace of 1s. I haven't been taught the integral method of inverse. Only the formula based , splitting terms kind of thing. I used matlab and found it was dirac delta. But how do I get to it without using the integral for inverse?
  2. jcsd
  3. Feb 6, 2013 #2
    When I look at this in Mathematica I get a derivative of the delta function, in other words:
    \mathcal{L}^{-1}\left\{s\right\} = \frac{d}{dt}\delta(t)
  4. Feb 6, 2013 #3
    Hey yeah, figured that out. It's actually by this property:

    inverse( df/dt)= s F(s) where laplace(f) = F(s)

    laplace ( dirac delta )=1 ( known property )

    laplace( d(diracdelta)/dt ) = s*(1)

    hence, inverse( s) = d(diracdelta)/dt
  5. Feb 6, 2013 #4

    Almost, be careful that:

    [itex]\mathcal{L}\left\{ \frac{\mathrm{d}f}{\mathrm{dt}} \left(t\right)\right\}=s\mathcal{L}\left\{f\left(t\right)\right\}[/itex]

    Not the inverse as you mentioned it.
  6. Feb 6, 2013 #5
    yeah yeah. That's what I meant. Typo.
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