I have a unit(or Heaviside) step function in discrete form:
$$\text u[n]=\begin{cases} 0, & n < 0, \\1, & n \ge 0, \end{cases}$$
and in continuous form:
$$\text u(t)=\begin{cases} 0, & t < 0, \\1, & t \ge 0, \end{cases}$$
It's ambigious I understand but let's ignore the zero argument.
Now we know the condition for a function to be periodic as it should repeat itself over $-\infty$ to $\infty$. If I use that condition the above signal is clearly a aperiodic but here is the problem, I have seen that people seem to have a two-sided opinion about the periodicity of the step function:
- Some say it's aperiodic using the above condition.
- Some say it's periodic because after ignoring the
zeros ($t<0$ or $n<0$), it's is periodic.
I strongly feel that the second set of people are wrong. I therefore would like to confirm my understanding with you people.
Best Answer
Definition 1. A function $f:D\to\mathbb{R}$, $D\subseteq\mathbb{R}$ is said to be periodic if there exists a positive real number $\omega$ such that $f(t)=f(t+\omega)$ for all $t\in D$. The number $\omega$ is called the period of $f$.
This very basic definition can be found in (almost) every text of real analysis, and is in favor of the first of the two ways of thinking you describe in your question: it does not exclude zeros from the range of the function $f$, otherwise it would exclude even the important mathematical constant $f\equiv 0$. The subtlety on which probably rely the supporters of the second way of thinking is that, by ruling out zeros in the range, you are implicitly considering a restriction of $H(x)$ to a subdomain where it is periodic according to definition 1. However, $H(x)$ is globally defined on $\mathbb{R}$ so if you state it has a property tout court, it must hold on all its domain of definition: in sum, according to definition 1, $H(x)$ is non-periodic.