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Inverse laplace of 1s

by indianaronald
Tags: inverse laplace
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indianaronald
#1
Feb5-13, 10:09 PM
P: 21
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?
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16180339887
#2
Feb6-13, 12:52 AM
16180339887's Avatar
P: 38
When I look at this in Mathematica I get a derivative of the delta function, in other words:
[tex]
\mathcal{L}^{-1}\left\{s\right\} = \frac{d}{dt}\delta(t)
[/tex]
indianaronald
#3
Feb6-13, 01:33 AM
P: 21
Quote Quote by 16180339887 View Post
When I look at this in Mathematica I get a derivative of the delta function, in other words:
[tex]
\mathcal{L}^{-1}\left\{s\right\} = \frac{d}{dt}\delta(t)
[/tex]
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

jfgobin
#4
Feb6-13, 01:19 PM
P: 90
Inverse laplace of 1s

Indianaronald,

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.
indianaronald
#5
Feb6-13, 10:07 PM
P: 21
Quote Quote by jfgobin View Post
Indianaronald,

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.
yeah yeah. That's what I meant. Typo.


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