# Does $y[n] = x[n] \star (u[n]-u[n-2])$ have memory or is it memoryless?

$$y[n] = x[n] \star (u[n]-u[n-2])$$, 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.

• Absolutely right, I already updated my post – dprozz122 Jun 18 '19 at 17:16
• 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? – Matt L. Jun 18 '19 at 18:11
• Yes, that is right. – dprozz122 Jun 18 '19 at 18:13
• me, the Notation Nazi, didn't even notice that, Fat. thanks for pointing it out. – robert bristow-johnson Jun 18 '19 at 19:23
• but i did notice a small mistake with an inequality that i fixed. – 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.