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In LTI systems there are two famous functions the unit step function and the unit impulse functions. And they are defined as follows. $$ u[n] = \begin{cases} 1, & \text{if $n$ $\ge$ 0} \\ 0, & \text{if $n$ $\lt$ 0} \end{cases} $$ $$ \delta[n] = \begin{cases} 1, & \text{if $n$ $=$ 0} \\ 0, & \text{if n $\neq$ 0}. \end{cases} $$

They also define $u[n]$ as $$\sum_{k=-\infty}^{n} \delta[k]$$

Isn't this something like $ ... + \delta[-3] + \delta[-2] + \delta[-1] + \delta[0] + \delta[1] + \delta[2] + \delta[3] + ... + \delta[n]$ that gives just one(1)? Because $\delta[n] = 1$ when $n = 0$. How can we say it is a step function?

Davide Giraudo
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Padmal
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    $u[-1] = ... + \delta[-2] + \delta[-1] = ...+0+0 = 0$, $u[17] = ... + \delta[-1] + \delta[0] + \delta[1] + ... + \delta[17] = ...+0+1+0+...+0 = 1$. – Abstraction Dec 25 '15 at 14:40
  • Aha now I'm getting it. $u[n]$ doesn't mean a continuous line but a point only. So it makes sense. :) – Padmal Dec 25 '15 at 14:54

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