DT unit step; periodicity of e^(jωn)

Discrete-time unit step signal

The discrete-time unit step $u[n]$ is a sequence that is zero for negative $n$ and one for $n \ge 0$:

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

It can be written as a sum of shifted impulses, $u[n] = \sum_{k=0}^{\infty} \delta[n-k]$, and it is used to make signals causal (switch them on at $n = 0$).

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

Condition for $x[n] = e^{j\omega n}$ to be periodic

$x[n]$ is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$:

$$\begin{aligned} e^{j\omega (n+N)} &= e^{j\omega n} \\ e^{j\omega n}\, e^{j\omega N} &= e^{j\omega n} \\ \Rightarrow e^{j\omega N} &= 1 \end{aligned}$$

$e^{j\theta} = 1$ only when $\theta$ is an integer multiple of $2\pi$, so

$$\omega N = 2\pi m \quad \Rightarrow \quad \frac{\omega}{2\pi} = \frac{m}{N}, \quad m, N \text{ integers}$$

Condition: $e^{j\omega n}$ is periodic only if $\omega/2\pi$ is a rational number. The fundamental period is the smallest such $N$: write $\omega/2\pi = m/N$ in lowest terms, then $N$ is the period.

Example: $\omega = 3\pi/4$ gives $\omega/2\pi = 3/8$, so $N = 8$. But $\omega = 1/6$ gives $\omega/2\pi = 1/(12\pi)$, which is irrational, so $e^{jn/6}$ is not periodic (unlike the continuous-time $e^{j\omega t}$, which is periodic for every $\omega$).