## Inverse laplace of 1s

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?

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 When I look at this in Mathematica I get a derivative of the delta function, in other words: $$\mathcal{L}^{-1}\left\{s\right\} = \frac{d}{dt}\delta(t)$$

 Quote by 16180339887 When I look at this in Mathematica I get a derivative of the delta function, in other words: $$\mathcal{L}^{-1}\left\{s\right\} = \frac{d}{dt}\delta(t)$$
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

## Inverse laplace of 1s

Indianaronald,

Almost, be careful that:

$\mathcal{L}\left\{ \frac{\mathrm{d}f}{\mathrm{dt}} \left(t\right)\right\}=s\mathcal{L}\left\{f\left(t\right)\right\}$

Not the inverse as you mentioned it.

 Quote by jfgobin Indianaronald, Almost, be careful that: $\mathcal{L}\left\{ \frac{\mathrm{d}f}{\mathrm{dt}} \left(t\right)\right\}=s\mathcal{L}\left\{f\left(t\right)\right\}$ Not the inverse as you mentioned it.
yeah yeah. That's what I meant. Typo.

 Tags inverse laplace