How to find Laplace transform of periodic function

I have function $f(t) = - \frac{1}{4}(t-2)^2+1$ with a period $T=4$. Pic

I know that in order to find the Laplace transform of the periodic function we apply the following rule: $F(s) = \frac{1}{1-e^{sT}} \int_{0}^{T}e^{-sT}f(t) dt$.

But how can I use this for my task?

Solution 1:

Let $\tilde f(t)$ denote the "periodic extension" of $f(t)$. Then we can write

$$\tilde f(t) = \sum_{k=0}^\infty f(t-4k) \big(\theta(t-4k) - \theta(t-4(k+1))\big)$$

where $\theta$ denotes the step/Heaviside theta function,

$$\theta(t) = \begin{cases} 1 & \text{if } t \ge 0 \\ 0 & \text{if } t < 0 \end{cases}$$

The Laplace transform of $\tilde f$ is then

$$\mathscr L_s\left\{\tilde f(t)\right\} = \int_0^\infty \tilde f(t) e^{-st} \, dt = \sum_{k=0}^\infty \int_{4k}^{4(k+1)} f(t-4k) e^{-st} \, dt$$

because for a given integer $k$, we have $\theta(t-4k)-\theta(t-4(k+1))=1$ whenever $t$ is between the consecutive multiples of $4$ ($4k$ and $4(k+1)$), and $0$ elsewhere.

Substituting $u=t-4k$ shows that

$$\int_{4k}^{4(k+1)} f(t - 4k) e^{-st} \, dt = \int_0^4 f(u) e^{-s(u+4k)} \, du = e^{-4ks} \int_0^4 f(u) e^{-su} \, du$$

and hence

$$\begin{align} \mathscr L_s \left\{\tilde f(t)\right\} &= \sum_{k=0}^\infty e^{-4ks} \int_0^4 f(u) e^{-su} \, du \\[1ex] & = \sum_{k=0}^\infty e^{-4ks} \frac{2s-1+(2s+1)e^{-4s}}{2s^3} \\[1ex] & = \frac{2s-1+(2s+1)e^{-4s}}{2s^3} \sum_{k=0}^\infty e^{-4ks} \\[1ex] & = \frac{2s-1+(2s+1)e^{-4s}}{2s^3} \cdot \frac{1}{1-e^{-4s}} \\[1ex] & = \frac{1}{1-e^{4s}} \cdot \frac{(1-2s)e^{4s}-2s-1}{2s^3} \end{align}$$