Find the Laplace transform?

Question

Find the Laplace transform of $f(t) = 1 + (1 - t)u_1(t) + (t-2)u_3(t)$.

Obviously each term of the function must be of $f(t - c)u_c(t)$ or be clearly transformable. Thus we have for our first term simply $\frac{1}{s}$. Our second term has been shifted by $f(t) = -t \to f(t - 1) = 1 - t$. Our third term has been shifted by $f(t) = t + 1$ and so $f(t - 3) = t - 2$.

Now I have to transform it. I get $$G(t) = \frac{1}{s} + e^{-s}(t - \frac{t^2}{2}) + e^{-3s}(\frac{t^2}{2} - 2t)$$

I don't know if this is correct. I used transform #28 from the following table : http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

Answer 1

$f(t) = 1+u_1(t) - t u_1(t) + t u_3(t) - 2 u_3(t)$.

You need to take care with multiplication by $t$. The rule is ${\cal L} (t \mapsto t f(t))(s) = - \frac{d \hat{f}(s)}{ds} $.

Then (abusing notation slightly) \begin{eqnarray} \hat{f}(s) &=& \frac{1}{s} + \frac{1}{s} e^{-s} - (- \frac{d}{ds} (\frac{1}{s} e^{-s})) + (- \frac{d}{ds} (\frac{1}{s} e^{-3s}))-2\frac{1}{s} e^{-3s} \\ &=& \frac{1}{s} + \frac{1}{s} e^{-s} - \frac{(s+1)}{s^2} e^{-s} + \frac{3s+1}{s^2} e^{-3s} -2\frac{1}{s} e^{-3s} \\ &=& \frac{1}{s} -\frac{1}{s^2}e^{-s} + \frac{(s+1)}{s^2} e^{-3s} \end{eqnarray}

When you apply the transform, all '$t$'s should disappear.