Example 1
Consider the function
\[f(t) = \left\{\begin{array}{ll} 5 &: 0\le t< 2\\ \pi &: 2\le t< 3\\ -4 &: 3\le t< 2\pi\\ 2 &: t\ge 2\pi \end{array}\right..\]Here is a graph of $f$:
I’ve included different colors in the graph so that we can discuss the different parts of the function.
How do we handle each different part?
We recall that $u_a-u_b$ is the function which is 1 for $a \le t< b$ and $0$ for all other $t$’s.
So the red part of the graph (i.e. what happens for $0\le t < 2$) can be described as turning on the constant function $5$ for the times between 0 and 2. That is
\[5(u_0 - u_2) \quad \text{is the red part of the graph}.\]
Similarly,
\[\pi (u_2-u_3) \qquad\text{is the light blue part of the graph,}\\ -4(u_3-u_{2\pi}) \qquad \text{ is the purple part of the graph,}\\ 2 u_{2\pi} \qquad \text{is the black part of the graph.}\]Adding all of these together gives the decomposition of $f$ into step functions:
\[f(t) = 5(u_0-u_2) + \pi (u_2-u_3) - 4 (u_3-u_{2\pi}) + 2u_{2\pi}\\ = 5u_0 + (\pi-5) u_2 + (-4-\pi)u_3 + 6 u_{2\pi}.\]Typically we write $u_0 = 1$, since we usually just care about positive time $t\ge 0$.
\[f(t) = 5+(\pi-5) u_2 + (-4-\pi)u_3 + 6 u_{2\pi}.\]Example 2
Consider the function
\[f(t) = \left\{ \begin{array}{ll} t^2 &:0\le t<2\\ 4 &: 2\le t < 5\\ 4 - (t-5)^2 &: t\ge 5 \end{array}\right..\]Here is the graph of $f$
Again, we think of $g(t) (u_a-u_b)$ as the function which turns on the function $g(t)$ only during the time $a\le t<b$.
Therefore,
\(t^2(u_0-u_2) \quad \text{is the red part,}\\ 4(u_2-u_5) \quad \text{is the light blue part,}\\ \left(4-(t-5)^2\right)(u_5) \quad \text{is the green part.}\) Adding these together gives the decomposition of $f$.
\[f(t) = t^2(u_0-u_2) +4(u_2-u_5) + (4-(t-5)^2)u_5\\ = t^2 u_0 + (4-t^2)u_2 + (-(t-5)^2) u_5\\ =t^2 +(4-t^2)u_2 - (t-5)^2 u_5.\]