Explain then answer if you can?

Question

What is the inverse LaPlace Transform of 7[e^(-7s)]/(s+7) I suspect it is the result of two functions one of whome is right time shifted by 7 sec.

Answer 1

Let f(t) be an arbitrary function.
Its Laplace transform is defined by
L{f(t} =
∞
∫ f(t) · e^(-s·t) dt
0

You can shift it to the right by a heavy side step function
u(t-t₀) = 0 for t u(t-t₀) = 1 for t≥t₀
The Laplace transform of the shifted function f(t - t₀) = u(t-t₀)·f(t) is
L{f(t-t₀} =
∞
∫ u(t-t₀)·f(t) · e^(-s·t) dt =
0
∞
∫ f(t) · e^(-s·t) dt =
t₀
(substitute t' = t - t₀)
∞
∫ f(t) · e^(-s·(t'+ t₀) dt' =
0
∞
∫ f(t) · e^(-s·(t'+ t₀) dt' · e^(-s·t₀)=
0
e^(-s·t₀) · L{f(t'}


To find the inevserse transform, first find the inverse transform
of the terms after the exponential function
L{f(t'} -----> f(t')
Then replace t' by t - t₀. That's it.

For the given Laplace transform
L{f(t} = e^(-7s) ·7 / (s +7)
The inverse transform of
L{f(t'} = 7/ (s +7)
can be found from the table of Laplace transforms
f(t') = 7·e^(-7t')
with t' = t -7
f(t) = 7·e^(-7·(t -7) ) = 7·e^(49 -7t)
voila

Answer 2

oops! i've ran out of charecters to work with.