- #1
quietrain
- 655
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Homework Statement
is it possible to find the laplace transform of f(t-b) ? i don't know if its possible, i am just trying.
The Attempt at a Solution
so, where integral from 0 to infinity,
[itex]\int[/itex] f(t-b) e-stdt
let t-b = z
=[itex]\int[/itex]f(z) e-s(z+b)dz
=[itex]\int[/itex]f(z) e-sze-sbdz
=e-sb f(s) , where f(s) is the laplace transform?
if this is correct, then
question) is f(s) the laplace transform of f(z)? i.e of f(t-b) ? OR is it the laplace transform of f(t) only?
because, from above, the expressions are telling me that f(s) is the laplace transform of f(z) which is f(t-b)
but if that is the case, then it wouldn't make any sense to even compute the laplace transform by the above method. because it would then mean L { f(t-b) } = f (s) ?
if the entire above is wrong, then how do i compute it?
because i know the inverse laplace transform of e-bsf(s) is f(t-b), so the reverse has to be true