"Inverse" of a step function How can I write the function below
$$f(x)=\left\{\begin{array}{ll} 1, & 0\leq t\leq 1, \\ 0, & t>1 \end{array}\right.$$
using the unit step function?
I mean, I don't know how could I write $f$ like:
$$f(x)=u(t)+g(t)$$
for some function $g$.
 A: In general, if you have a step function such as
$$f(t)=\begin{cases}g_0(t)&0\le t_0\\
                    g_1(t)&t_0\le t_1\\
                    g_2(t)&t_1\le t
       \end{cases}$$
It can be rewritten in terms of the step functions as follows
$$ f(t)=g_0(t)+[g_1(t)-g_0(t)]u(t-t_0)+[g_2(t)-g_1(t)]u(t-t_1) $$
For example, suppose we have
$$f(t)=\begin{cases}2t&0\le 1\\
                    2&1\le 3\\
                    8-2t&3\le t<4\\
                    0&4\le t
       \end{cases}$$
with the following graph:

This can be re-written as
\begin{eqnarray}
f(t)&=&2t+(2-2t)u(t-1)+[(8-2t)-2]u(t-3)+[0-(8-2t)]u(t-4)\\
&=&2t-2(t-1)u(t-1)-2(t-3)u(t-3)+2(t-4)u(t-4)
\end{eqnarray}
In a case such as this example where $f(t)$ is a continuous function one will always find that it is perfectly set up for finding the Laplace transform using
$$ \mathcal{L}\{g(t-a)u(t-a)\}=\mathcal{L}\{g(t)\}e^{-as} $$
If $f$ is not a continuous function one can use
$$ \mathcal{L}\{g(t)u(t-a)\}=\mathcal{L}\{g(t+a)\}e^{-as} $$
