How to rewrite a piecewise function in terms of the Heaviside function

Question

Let's say I have a piecewise function:

$$f(x) = \begin{cases}x ,& 0 \leq x \leq 1 \\1 ,& 1 \leq x\end{cases}$$

How can I rewrite this in terms of the Heaviside function $u(x-a)$?

Answer 1

I gave a good rundown of a mechanical method here. The main idea is to change everything to Iverson brackets before finally switching to the unit step function, since there is the relationship

$$[x\geq a]=[x-a\geq0]=\begin{cases}1&\text{if }x-a\geq0\\0&\text{if }x-a<0\end{cases}=u(x-a)$$

With

$$f(x) = \begin{cases} x & 0 \leq x < 1 \\ 1 & 1 \leq x \end{cases}$$

(and assuming that the function is zero in all other cases), translation to the Iverson convention is easy:

$$f(x)=x[0 \leq x < 1]+[1 \leq x]$$

and we can then do some massaging:

$$\begin{align*} f(x)&=x[0 \leq x < 1]+[1\leq x]\\ &=x[x \geq 0][x < 1]+[x-1 \geq 0]\\ &=x[x \geq 0][\lnot(x \geq 1)]+[x-1 \geq 0]\\ &=x[x \geq 0](1-[x \geq 1])+[x-1 \geq 0]\\ &=x[x \geq 0](1-[x-1 \geq 0])+[x-1 \geq 0]\\ &=x[x \geq 0]-x[x \geq 0][x-1 \geq 0]+[x-1 \geq 0]\\ &=x\,u(x)-x\,u(x)u(x-1)+u(x-1)\\ \end{align*}$$

where the properties $[p\text{ and }q]=[p][q]$ and $[\lnot p]=1-[p]$ of the Iverson bracket were useful.

One can do a further simplification, since $u(x)u(x-1)=[(x \geq 0)\text{ and }(x \geq 1)]=[x \geq 1]=u(x-1)$. We then finally have

$$f(x)=x\,u(x)-x\,u(x-1)+u(x-1)=x\,u(x)+(1-x)u(x-1)$$

Answer 2

At $x=0$ the slope grows of $1$ adding a term $x\cdot u(x-0)$
At $x=1$ the slope change ends subtracting a term $(x-1)\cdot u(x-1)$

So that $f(x)=x\cdot (u(x-0)-u(x-1))+u(x-1)$.

Answer 3

Given the pieces your function is made of, you want to look for functions g(x) and h(x) such that f(x)=g(x)u(x)+h(x)u(x-1). I let you determine the functions g and h.

Answer 4

In general, assuming that the conditions $c_1, \ldots, c_n$ are mutually exclusive, the translation would be $$ \begin{eqnarray*} f(x) & = & \begin{cases} f_1(x), & c_1 \\ \ldots \\ f_n(x), & c_n \\ \end{cases} \\ & = & f_1(x) \cdot \mathcal{U}(c_1) + \ldots + f_n(x) \cdot \mathcal{U}(c_n) \end{eqnarray*} $$ where $\mathcal{U}(c_n)$ is the encoding of $n$th condition.

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Relevant: Superposition Principle used a lot in engineering.