Explain then answer if you can?

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<t₀

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

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