Proof of L-1{F(s)} = L-1{F(z-2)} = g(t) * e-2t

[yitriana]

to calculate the inverse laplace transform of a function F(s), s+2 was replaced with z for convenience. the inverse laplace transform of z was found--let's denote the function g(t).

now, how do i prove that L^-1{F(s)} = L^-1{F(z-2)} = g(t) * e^-2t

i am attempting to prove by rewriting L^-1{F(z-2)} in terms of L^-1{F(z)*something} but i don't know what exactly to do.

this is not a homework question, just attempt at a proof.

 

[pbandjay]

I find it easiest to just show the shifting property using the integral definition and then using that information to show the inverse Laplace transform.

$$F(s-a) = \int_0^\infty{e^{-(s-a)t}f(t)dt} = \int_0^\infty{e^{at}e^{-st}f(t)dt} = L[e^{at}f(t)]$$

And therefore $$L^{-1}[F(s-a)] = e^{at}f(t) = e^{at}L^{-1}[F(s)]$$

I am not sure if you were looking for this or something else in particular. It's 5am here and I'm having some trouble seeing straight.. :shy: