Step Function (heaviside) laplace xform

[chota]

Hi, I currently have this problem to solve and I can't seem to figure it out

it goes like this


Find the Laplace transform of the given function:

f(t) = { 0 t<2, (t-2)^2 t>=2

I tried working it out and this is where i get stuck

f(t) = (t-2)^2 * u(t-2)

I am not sure if I got the write function for f(t), but if I did, I am not sure how to go on with solving this.
Any help is appreciated, Thank YOu
Chota

 

[HallsofIvy]

Did you consider integrating it directly?
The laplace transform of function f(t) is defined as
\int_0^\infty e^{-st}f(t)dt[/itex]<br /> and here<br /> \int_0^\infty e^{-st}(t-2)^2 u(t-2)dt= \int_2^\infty (t-2)^2e^{-st}dt<br /> <br /> and that can be integrated using integration by parts, twice.

 

[coomast]

Hi, I currently have this problem to solve and I can't seem to figure it out

it goes like this


Find the Laplace transform of the given function:

f(t) = { 0 t<2, (t-2)^2 t>=2

I tried working it out and this is where i get stuck

f(t) = (t-2)^2 * u(t-2)

I am not sure if I got the write function for f(t), but if I did, I am not sure how to go on with solving this.
Any help is appreciated, Thank YOu
Chota

You need to make use of a translation property of the Laplace transform. It states that:
if F(s) is the Laplacetransform of f(t) then L{f(t-a)u(t-a)}=exp(-as)F(s)

Can you apply this to your problem?