I have a diagram that looks like below:
First, I have attempted to express the diagram function regarding f(t) as below:
\begin{equation} f(t) = \begin{cases} t & \text{0 < t ≦ 1}\\ 2-t & \text{1 < t ≦ 3}\\ t-4 & \text{3 < t ≦ 5}\\ 6-t & \text{5 < t ≦ 7}\\ t-8 & \text{7 < t ≦ 8}\\ 0 & \text{t < 0, 8 < t} \end{cases} \end{equation}
I am not certain about this, but I think it should make sense a little bit if we put the equation to the corresponding interval. Now I would like to find the following Laplace Transform based on the interval respectively:
\begin{equation} f_1(t) = \begin{cases} t & \text{0 < t ≦ 1}\\ 2-t & \text{1 < t ≦ 2}\\ 0 & \text{t < 0, t > 2} \end{cases} \end{equation}
\begin{equation} f_2(t) = \begin{cases} 2-t & \text{2 < t ≦ 3}\\ t-4 & \text{3 < t ≦ 4}\\ 0 & \text{t ≦ 2, t > 4} \end{cases} \end{equation}
\begin{equation} f(t) = \begin{cases} t & \text{0 < t ≦ 1}\\ 2-t & \text{1 < t ≦ 3}\\ t-4 & \text{3 < t ≦ 5}\\ 6-t & \text{5 < t ≦ 7}\\ t-8 & \text{7 < t ≦ 8}\\ 0 & \text{t < 0, 8 < t} \end{cases} \end{equation} (Note: the above equations are derived from the diagram that I see. Only the intervals are given in the case)
Based on the definition of Laplace Transform and its common formula sheet, the first Laplace transform I wrote like this:
\begin{equation} F_1(s) = \begin{cases} \mathscr{L}\{t\}=\int_{0}^{\infty}te^{-st}dt = \dfrac{1}{s^2} & \text{0 ≦ t ≦ 1}\\ \mathscr{L}\{2-t\}=\int_{0}^{\infty}(2-t)e^{-st}dt = \dfrac{2}{s} - \dfrac{1}{s^2} & \text{1 < t ≦ 2}\\ 0 & \text{t < 0, t > 2} \end{cases} \end{equation}
However, my question is, does it make sense to simply add the equation above as the result of the Laplace transform like this:
\begin{equation} F_1(s) = \dfrac{2}{s} \end{equation}
If this is incorrect, what could be the answer?
Please help me with this. Thank you for your advice and time!
Defining
$$ \Lambda(t) = \cases{t, \ \ \ \ \ \ \ \ \ 0\lt t\le 1\\ 2-t, \ \ 1\lt t\le 2\\ 0, \ \ \ \ \ \ \ \ \ \text{otherwise} } $$
we have
$$ f(t) = \Lambda(t)-\Lambda(t-T)+\Lambda(t-2T)-\Lambda(t-3T) $$
and thus
$$ \hat f(s) = (1-e^{-2s}+e^{-4s}-e^{-6s})\hat \Lambda(s) $$
with
$$ \hat \Lambda(s) = \frac{e^{-2 s} \left(e^s-1\right)^2}{s^2} $$
NOTE
$$ \Lambda(t) = t\left(\theta(t)-\theta(t-1)\right)+(2-t)\left(\theta(t-1)-\theta(t-2)\right) $$
where $\theta(t)$ is the heaviside unit step function. Here
$$ \int_0^{\infty} t\left(\theta(t)-\theta(t-1)\right)e^{-s t}dt = \int_0^{1} t e^{-s t}dt = \frac{1-e^{-s} (s+1)}{s^2} $$
etc. Also
$$ \int_0^{\infty}f(t-T)e^{-s t}dt = e^{-sT}\hat f(s) $$