# Very simple question about signal periodicity

$$x[n] = u[n]+u[-n]$$

Is it periodic or not? My answer is

$$u[n] = {1 , n\geqslant0}$$
$$u[-n] = {1 , n\leqslant0}$$

which means that the signal $x[n]$ is always equal to $1$ from $-\infty$ to $+\infty$, but it equals $2$ at $n=0$. Am I right?

• It depends on the definition of $u[n]$. Both $u=1$ and $u=1/2$ are common conventions.
– MBaz
May 23, 2016 at 21:56
• What if we assumed that $u = 1$ ? May 23, 2016 at 22:01
• If $u=1$, then yes, you are right. The signal is always $1$ except at $n=0$, where it takes the value of $2$. May 24, 2016 at 0:09

For the step function $u[n]$ defined as $$u[n]=\begin{cases}1\text{ n\geq0,}\\ 0\text{ otherwise,}\end{cases}$$ the function $x[n]=u[n]+u[-n]$ is given by $$x[n]=\begin{cases}2\text{ n=0,}\\1\text{ otherwise.}\end{cases}$$ Clearly the signal $x[n]$ is not periodic.
no, but $x[n]$ is, what we call, an "even-symmetry" function:
$$x[-n] = x[n] \quad \forall n \in \mathbb{Z} \ .$$
what makes $x[n]$ periodic is $$x[n+N] = x[n] \quad \forall n \in \mathbb{Z}$$ which is really the premise for the DFT of length $N$.