Laplace transformation using second shifting theorem

Question

$$f(t)=\begin{cases}t^2,&t\lt4\\t,&t\geqslant4\end{cases}.$$

can anyone tell me how to evaluate the solution? I really get stuck.

Answer 1

To use the second shifting theorem, you need to put it into the appropriate form i.e. $g(t-c)\cdot u(t-c)$. Well technically you only really need to put part of $f(t)$ in the appropriate form.

To start off, note that your function can be described in terms of the heaviside function $u(t)$ as

$$\begin{aligned}f(t) &=t^2-t^2\cdot u(t-4)+t\cdot u(t-4)\\ &= t^2-(t^2-t)\cdot u(t-4)\end{aligned}$$

Not quite the form we need for the second shifting theorem, but we can apply the good ol' "add zero" trick:

$$\begin{aligned}f(t) &= t^2-(t^2-t)\cdot u(t-4)+(7t-16)\cdot u(t-4)-(7t-16)\cdot u(t-4)\\ &=t^2-(t^2-8t+16)\cdot u(t-4)-7t\cdot u(t-4)+16\cdot u(t-4)\\ &=t^2-(t-4)^2\cdot u(t-4)-7(t-4)\cdot u(t-4)-12\cdot u(t-4)\end{aligned}$$

When applying the Laplace transform to this expression, we can apply the transformation for each individual term (by linearity).

Oh look, what do we have for individual terms?

$t^2$: Super easy to transform, you and I can do this in our sleep.

$-(t-4)^2\cdot u(t-4)$: Of the form $g(t-4)\cdot u(t-4)$ where $g(t)=-t^2$. Use the second shifting theorem on this bad boy.

$-7(t-4)\cdot u(t-4)$: Of the form $g(t-4)\cdot u(t-4)$ where $g(t)=-7t$. Apply the second shifting theorem here as well.

$-12\cdot u(t-4)$: Standard transformation, either from memory or by consultation of the holy table of Laplace transforms.

Good luck!