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{u_c 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 ?!

 

[BvU]

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

 

[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!

 

[Ray Vickson]

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{u_c 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 $\text{answer} = e^{-2s} F(s)$.