The function f(t) is defined as follows: $\displaystyle f(t) = \begin{cases} t(t+1) & \ 0 < t < 2 \\ 1 & \ 2 \leq t \end{cases} $ Find $\displaystyle \mathcal{L} (f(t)) $.
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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
Originally Posted by MrJack1990 The function f(t) is defined as follows: $\displaystyle f(t) = \begin{cases} t(t+1) & \ 0 < t < 2 \\ 1 & \ 2 \leq t \end{cases} $ Find $\displaystyle \mathcal{L} (f(t)) $. If you are good at integration by parts, you can also do this from the definition of the LT: $\displaystyle \mathcal{L}f (s)=\int_0^{\infty} f(x) e^{-st}\; dt= \int_0^{2} t(t+1) e^{-st}\; dt + \int_2^{\infty} e^{-st}\; dt$ CB
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