Consider $ u(t-a) = \begin{cases} 0, & \text{if }t<a \\ 1, & \text{if }t\geq a \end{cases} $

How can we rewrite a function like $ f(t) = \begin{cases} \cos2t, & \text{if }0\leq t \lt 2\pi \\ 0, & \text{if }t\geq 2\pi \end{cases} $ in terms of the unit step function? My textbook writes this particular example as $f(t) = [1-u(t-2\pi)]\cos2t$, but I don't understand how this was formulated nor how I can formulate other piecewise functions in terms of the unit step similarly.


1 Answer 1


Consider the function $f(a,b) = u(t-a)-u(t-b); a>b$. This function is 1 in $[a,b)$ and 0 elsewhere. So, suppose you want to write $$g(x) = \begin{cases} g_1(x), & \text{if } a_1<x<a_2 \\ g_2(x), & \text{if } a_2<x<a_3\\ \ldots \end{cases}$$

Then, $$g(x) = g_1(x)f(a_2,a_1) + g_2(x)f(a_3,a_2)+\ldots$$

  • $\begingroup$ I was looking for a "formula" such this one for quite a while. But, I must correct you: You should change it to $f(a,b)=u(t-b)-u(t-a)$ $\endgroup$
    – E Be
    Commented Sep 16, 2014 at 16:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .