Question 5.4.6: Find the Laplace transform of f (t ), where f (t ) is the pi...

From the graph, we see that f has slope 1 on [0, 2) and slope −2 on [2, 3). Therefore, f can be defined piecewise by the rule

f(t)=\begin{cases}t, & if 0 ≤ t < 2 \\6-2t, & if 2 ≤ t < 3 \\0, & if 3 ≤ t \end{cases}

Using step functions, we can write f according to the formula

f (t ) = t [u(t )−u(t −2)]+(6−2t )[u(t −2)−u(t −3)]
= tu(t )+(6−3t )u(t −2)−(6−2t )u(t −3)

Applying the second shifting property, linearity, and familiar transforms, we see that

L[f (t )] = L[tu(t )]+L[(6−3t )u(t −2)]−L[(6−2t )u(t −3)]

= L[t]+e^{−2s} L[6−3(t +2)]−e^{−3s} L[6−2(t +3)]

= L[t]+e^{−2s} L[−3t]−e^{−3s} L[−2t ]

= \frac {1}{s^{2}}− \frac {3}{s^{2}} e^{−2s} + \frac {2}{s^{2}}e^{−3s}