Unit step and delta: CT and DT

Continuous time

Unit step $u(t)$:

$$u(t) = \begin{cases} 1, & t > 0 \\ 0, & t < 0 \end{cases}$$

It is discontinuous at $t = 0$ (value there is undefined or taken as 1/2). It is used to switch signals on at $t = 0$.

Unit impulse (Dirac delta) $\delta(t)$: zero everywhere except at $t = 0$, with unit area:

$$\delta(t) = 0 \text{ for } t \neq 0, \qquad \int_{-\infty}^{\infty}\delta(t)\,dt = 1$$

It is the limit of a rectangular pulse of width $\Delta$ and height $1/\Delta$ as $\Delta \to 0$. Sifting property: $\int x(t)\delta(t-t_0)\,dt = x(t_0)$.

Discrete time

$$u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases} \qquad \delta[n] = \begin{cases} 1, & n = 0 \\ 0, & n \neq 0 \end{cases}$$

$\delta[n]$ is a simple sequence with value exactly 1 at $n = 0$ (no limiting process needed). Any sequence can be written as $x[n] = \sum_k x[k]\delta[n-k]$.

 u(t)              delta(t)
 1 +---------      ^ area = 1
   |               |
 --+--------> t  --+--------> t
   0               0

 u[n]
  1         o  o  o  o ...
            |  |  |  |
 -----o--o--+--+--+--+---> n
     -2 -1  0  1  2  3

 delta[n]
  1   o
      |
 --o--+--o--> n
  -1  0  1

Relationship

Continuous time:

$$u(t) = \int_{-\infty}^{t}\delta(\tau)\,d\tau, \qquad \delta(t) = \frac{du(t)}{dt}$$

The step is the running integral of the impulse; the impulse is the derivative of the step (the jump of height 1 at $t=0$).

Discrete time:

$$u[n] = \sum_{k=-\infty}^{n}\delta[k] = \sum_{k=0}^{\infty}\delta[n-k], \qquad \delta[n] = u[n] - u[n-1]$$

The step is the running sum of the impulse; the impulse is the first difference of the step.