# Given the function begin{cases}e^{-t}& text{if } 0leq t<2 0&text{if } 2leq tend{cases}Express f(t) in terms of the shifted unit step function u(t -a)F(t)

First order differential equations

Given the function $$\begin{cases}e^{-t}& \text{if } 0\leq t<2\\ 0&\text{if } 2\leq t\end{cases}$$
Express f(t) in terms of the shifted unit step function u(t -a)
F(t) - ?
Now find the Laplace transform F(s) of f(t)
F(s) - ?

2020-11-09

Step 1
Since f(t)=0 for $$2\leq t \text{ and } t<0$$ so create step function: $$u(t)-u(t-2)$$ which will give those zero values.
Thus, $$f(t)=e^{-t}[u(t)-u(t-2)] \text{ or } f(t)=e^{-t}u(t)-e^{-t}u(t-2)$$
Step 2
Calculate Laplace transform:
$$F(s)=\int_{-\infty}^\infty f(t)e^{-st}dt$$
$$=\int_{-\infty}^\infty(e^{-t}u(t)-e^{-t}u(t-2))e^{-st}dt$$
$$=\int_{-\infty}^\infty(e^{-t}u(t))e^{-st}dt - \int_{-\infty}^\infty(e^{-t}u(t-2))e^{-st}dt$$
$$=\int_0^\infty e^{-(s+1)t}dt - \int_2^\infty e^{-(s+1)t}dt$$
$$=\frac{1}{(s+1)}- \frac{e^{-2(s+1)}}{(s+1)}$$
$$=\frac{1-e^{-2(s+1)}}{(s+1)}$$
Step 3
Thus, Laplace transform is $$F(s)=\frac{1-e^{-2(s+1)}}{(s+1)}$$