CT vs DT signals; u[n] = Σδ[n−k]

CT vs DT signals

Show that $u[n] = \sum_{k=0}^{\infty}\delta[n-k]$

Expand the right-hand side:

$$\sum_{k=0}^{\infty}\delta[n-k] = \delta[n] + \delta[n-1] + \delta[n-2] + \cdots$$

$\delta[n-k] = 1$ only when $n = k$, otherwise 0.

• For $n < 0$: $n - k < 0$ for every $k \ge 0$, so every term is 0. Sum $= 0$.
• For $n \ge 0$: exactly one term, the one with $k = n$, equals 1; all others are 0. Sum $= 1$.

$$\sum_{k=0}^{\infty}\delta[n-k] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases} = u[n]$$

Hence proved. (Equivalently, putting $m = n-k$: $u[n] = \sum_{m=-\infty}^{n}\delta[m]$, the running sum of the impulse.) Graphically, the step is a train of unit impulses at $n = 0, 1, 2, \ldots$:

   o  o  o  o  o ...   = delta[n] + delta[n-1] + ...
   |  |  |  |  |
 --+--+--+--+--+--> n
   0  1  2  3  4