Laplace Transform - Step Function

1. Aug 5, 2007

twiztidmxcn

1. The problem statement, all variables and given/known data

What is the Laplace transform of f(t) = t^2 - 18 for 0 < t < 3 and f(t) = (t-3)^2 for t>3?

2. Relevant equations

Laplace Transforms

3. Work

Using Heaviside/step function, made equation into:

f(t) = t^2 - 18 + u(t-3)( (t-3)^2 - (t^2-18) )

Then, using Laplace transforms, found:

L{f(t)} = ( 2 / s^3 ) - ( 18 / s ) + e^(-3s)*L{27-6t}

I know that I have to put t in the L{27-6t} into either t-3 or t+3, but not sure which.

I went under the assumption that it t+3 would substitute for the t, so applying Laplace transforms I found:

L{27-6t} = L{27-6(t+3)} = L{27-6t-18} = L{9-6t} = 9/s - 6/s^2

Leaving me with the final answer:

L{f(t)} = 2/s^3 - 18/s + e^-3s*(9/s - 6/s^2)

...any chance this is close to right?

Last edited: Aug 5, 2007
2. Aug 6, 2007

Cenarius_lo_ol_

em,you're quite right
Let
u(t-3)( (t-3)^2 - (t^2-18) )=u(t-3)g(t-3)
You're using the property of time shifting. Put t into t+3 and you get u(t)g(t).
Then
L{u(t-3)g(t-3)}=e^(-3s) L{u(t)g(t))}

ps:
g(t-3)=27-6t
g(t)=9-6t

Last edited: Aug 6, 2007