$y[n] = x[n] \star (u[n]-u[n-2])$, by its definition is has to be a system with memory since it is depended from a fraction of time in the past, but if we calculate the difference $u[n]-u[n-2]$, it results $u[n]$ for $0 \le n < 2$, which is memoryless. And that sign $\star$ is convolution.

  • $\begingroup$ Absolutely right, I already updated my post $\endgroup$ – dprozz122 Jun 18 '19 at 17:16
  • $\begingroup$ Your question is clearer now. So your system has an impulse response that is given by $h[n]=u[n]-u[n-2]$, where $u[n]$ is the unit step sequence, right? $\endgroup$ – Matt L. Jun 18 '19 at 18:11
  • $\begingroup$ Yes, that is right. $\endgroup$ – dprozz122 Jun 18 '19 at 18:13
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    $\begingroup$ me, the Notation Nazi, didn't even notice that, Fat. thanks for pointing it out. $\endgroup$ – robert bristow-johnson Jun 18 '19 at 19:23
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    $\begingroup$ but i did notice a small mistake with an inequality that i fixed. $\endgroup$ – robert bristow-johnson Jun 18 '19 at 19:25

The system $$y[n] = x[n] \star (u[n]-u[n-2])$$ where $u[n]$ is the unit step function, has memory.

Indeed the system is equivalent to $$y[n] = x[n] \star ( \delta[n] + \delta[n-1] ) \implies y[n] = x[n] + x[n-1]$$ and as it's clear from the given I/O relationship, the current value of the output $y[n]$, depends on the values input $x[n]$ at other times namely $x[n-1]$ (one step past value); hence the system has memory.

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