Laplace transform with time shift property

Question

ok so i have no idea how the inverse laplace went from $F(s)$ to $f(t)$. I understand $\frac{c}{s^2}$ => $ct$, and $\frac{b}{s}$ => $b$, but the $e^{-as}$ is what gets me. In my Laplace tables I know it has something to do with the time shift.

If you expand the first term of $F(s)$ you get \begin{equation} \frac{c}{s^2}-\frac{c}{s^2}e^{-as}. \end{equation} So in this case for $e^{-as}F(s)$, $F(s)$ would be $\frac{c}{s^2}$ correct? Then wouldn't the Laplace transform be $ct1(t-a)$? The time shift is confusing the hell out of me.

Answer 1

We have:

$$\mathscr{L}^{-1} (e^{-as}F(s)) = f(t-a)u(t-a)$$

• For $F(s) = \dfrac{1}{s^2}$, we would have $f(t) = t$.

Now, because of the $e^{-as}$ term, we have to apply the time-shift property to $f(t)$, by replacing $t = t-a$ using the above and get:

$$\mathscr{L}^{-1} \left(-c\dfrac{1}{s^2}e^{-a s}\right) = -c(t-a)u(t-a)$$

• For $c\dfrac{1}{s^2}$, we have $f(t) = t$, but no time shift, hence:

$$\mathscr{L}^{-1} \left(c\dfrac{1}{s^2}\right) = ct$$

• For $-b\dfrac{1}{s} e^{-as}$, we have $f(t) = 1$, and a time shift, hence:

$$\mathscr{L}^{-1} \left(-b\dfrac{1}{s} e^{-as}\right) = -bu(t-a)$$

Answer 2

$ct*1(t-a)$ is not the Laplace transform of $\frac{c}{s^2}e^{-as}$, because you haven't shift the function.
The function is $f(t)=t$, if you want to shift this function of a quantity $a$ you obtain: $f(t-a)=t-a$

In the second part the function is just $f(t)=1$, if you shift this function you obtain always $f(t-a)=1$, because is constant all over the axis $t$.