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

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?

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    $\begingroup$ It depends on the definition of $u[n]$. Both $u[0]=1$ and $u[0]=1/2$ are common conventions. $\endgroup$ – MBaz May 23 '16 at 21:56
  • $\begingroup$ What if we assumed that $u[0] = 1$ ? $\endgroup$ – Hossam Houssien May 23 '16 at 22:01
  • $\begingroup$ If $u[0]=1$, then yes, you are right. The signal is always $1$ except at $n=0$, where it takes the value of $2$. $\endgroup$ – Tendero May 24 '16 at 0:09
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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.

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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 $N$.

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