# L.Transform of Piecewise Function

• Aug 27th 2009, 08:11 AM
MrJack1990
L.Transform of Piecewise Function
The function f(t) is defined as follows:

$
f(t) = \begin{cases} t(t+1) & \ 0 < t < 2 \\ 1 & \ 2 \leq t \end{cases}$

Find $\mathcal{L} (f(t))$.
• Aug 27th 2009, 06:45 PM
Calculus26
Rewrite using the unit step function u(t)

f(t) = u(t)(t)(t-1) + u(t-2)[1- t(t-1)]

use L{u(t-a)f(t-a)} = e^(-as)F(s)

where F(s) = L{f(t)}

note 1-t(t-1) = -t^2+t +1 = -(t-2)^2 -4(t-2) - 3
• Aug 27th 2009, 08:31 PM
CaptainBlack
Quote:

Originally Posted by MrJack1990
The function f(t) is defined as follows:

$
f(t) = \begin{cases} t(t+1) & \ 0 < t < 2 \\ 1 & \ 2 \leq t \end{cases}$

Find $\mathcal{L} (f(t))$.

If you are good at integration by parts, you can also do this from the definition of the LT:

$\mathcal{L}f (s)=\int_0^{\infty} f(x) e^{-st}\; dt=
\int_0^{2} t(t+1) e^{-st}\; dt + \int_2^{\infty} e^{-st}\; dt$

CB