# Laplace Transform of this function

Pouyan

## Homework Statement

We want to find the Laplace transform for

f(t): 0 for t≤2 and (t-2)2 for t≥2

## Homework Equations

I know that Lap{uc f(t-c)} = e-csLap{f(t)}=e-csF(s)
I rewrite f(t)=0+g(t) where g(t) = 0 for 0≤t<2 and (t-2)2 for t≥2
so we can write f(t)=g(t)= u2(t)*(t-2)2
Lap{f(t)}= (e-2s/s) * Lap{(t-2)2}

## The Attempt at a Solution

The answer in my bok is 2*(e-2s)/(s3)
But Lap{(t-2)2} is (2/s3) - (4/s2)+(4/s)

why ?!

## Answers and Replies

Homework Helper
(t-2)^2 is not the same as your f(t) (especially in the range 0<t<2 ) !

• Pouyan
Pouyan
(t-2)^2 is not the same as your f(t) (especially in the range 0<t<2 ) !
TNX !

Now I've got it! Homework Helper
Dearly Missed

## Homework Statement

We want to find the Laplace transform for

f(t): 0 for t≤2 and (t-2)2 for t≥2

## Homework Equations

I know that Lap{uc f(t-c)} = e-csLap{f(t)}=e-csF(s)
I rewrite f(t)=0+g(t) where g(t) = 0 for 0≤t<2 and (t-2)2 for t≥2
so we can write f(t)=g(t)= u2(t)*(t-2)2
Lap{f(t)}= (e-2s/s) * Lap{(t-2)2}

## The Attempt at a Solution

The answer in my bok is 2*(e-2s)/(s3)
But Lap{(t-2)2} is (2/s3) - (4/s2)+(4/s)

why ?!

Start with your formula ##{\cal L}[u(t-c) f(t-c)](s) = e^{-cs}{\cal L}[f(t)](s)##. So if ##F(s) = {\cal L}[t^2](s)##, then the answer you want is ##\text{answer} = e^{-2s} F(s)##.