Remark. You proved in class that
L{f(t-a)u(t-a)}=e^(-as)L{f(t)}
This will be useful for two purposes: Finding Laplace transform of functions like
sin(t)u(t-\pi )
. have
sin(t)
and not
sin(t-\pi )
. But you can find a function si
f(t-\pi )=sin(t)
. Look at
7(b)
.) When taking inverse Laplace transforms on the right side wh functions look like
e^(-s)*(1)/(s^(2))
. Take inverse Laplace transform on both sides of equation (1):
f(t-a)u(t-a)=L^(-1){e^(-as)L{f(t)}}
Try to write
(1)/(s^(2))
as
L{*}
7. For the following find
f(t)
: (Going from
f(t-a)
to
f(t)
) (a)
f(t-1)=t^(3)+1
(b)
f(t-\pi )=sin(t)
(c)
f(t-2)=(t-2)^(2)+t