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A filter is defined as $ h \left[ n \right] = \delta \left[ n \right] - \delta \left[ n - 1 \right] $.

Given a signal $ h \left[ n \right] $ defined as:

$$ x \left [ n \right ] = \begin{cases} 1 & \text{ if } x \geq 0 \\ 0 & \text{ if } x < 0 \end{cases} $$

Let $ y \left[ n \right] = \left( x \ast h \right) \left[ n \right] $.
What is the value of $ y \left[ -1 \right], \, y \left[ 0 \right], \, y \left[ 1 \right], \, y \left[ 2 \right] $?

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closed as off-topic by Marcus Müller, Stanley Pawlukiewicz, lennon310, Dilip Sarwate, Matt L. Jul 17 '18 at 14:41

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    $\begingroup$ Where are you stuck? $\endgroup$ – Olli Niemitalo Jul 5 '18 at 9:23
  • $\begingroup$ Please use LaTeX for writing the questions. $\endgroup$ – Royi Jul 5 '18 at 9:56
  • $\begingroup$ Sorry sir, I dont know how to use latex so I am putting a screenshots so that I wont miss any data $\endgroup$ – praneeth bharath Jul 5 '18 at 12:59
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    $\begingroup$ I'm voting to close this question as off-topic because no effort shown $\endgroup$ – Stanley Pawlukiewicz Jul 5 '18 at 13:24
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The Discrete Delta Function, $ \delta \left[ n \right] $ is the identity operator of Linear Time Invariant Systems.

Moreover, since it LTI System we can computer for each element of the filter by itself.
So the first element of the filter, $ \delta \left[ n \right] $, just outputs the signal itself.
The other element $ \delta \left[ n - 1 \right] $ just shifts the input signal.
Since the input is 1 for any $ n \geq 0 $ we subtract 1 from 1 unless it is $ n = 0 $ then we subtract zero from 1.

Hence the solution is 0, 1, 0, 0.

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