DT periodicity; energy/power of e^(jω₀n)u[n]

Necessary condition for periodicity

A discrete-time signal is periodic with period $N$ (a positive integer) if $x[n+N] = x[n]$ for all $n$.

For $x[n] = e^{j\omega_0 n}$:

$$\begin{aligned} x[n+N] &= e^{j\omega_0 (n+N)} = e^{j\omega_0 n}\,e^{j\omega_0 N} \end{aligned}$$

For $x[n+N] = x[n]$ we need

$$e^{j\omega_0 N} = 1 \;\Rightarrow\; \omega_0 N = 2\pi m, \quad m \text{ an integer}$$ $$\frac{\omega_0}{2\pi} = \frac{m}{N} = \text{rational number}$$

So $e^{j\omega_0 n}$ is periodic only if $\omega_0/2\pi$ is a rational number. The fundamental period is the smallest positive integer $N = 2\pi m/\omega_0$, taking $m$ as the smallest integer that makes $N$ an integer (with $m$ and $N$ having no common factor). The same condition holds for $\cos(\omega_0 n+\theta)$ and $\sin(\omega_0 n+\theta)$, because they are sums of such exponentials.

This differs from continuous time, where $e^{j\omega_0 t}$ is periodic for every $\omega_0 \neq 0$ (period $2\pi/|\omega_0|$). In discrete time $n$ takes only integer values, so $N$ must be an integer.

Energy and power of $x[n] = e^{j\omega_0 n}u[n]$

$|x[n]|^2 = |e^{j\omega_0 n}|^2 u[n] = u[n]$, i.e. 1 for $n \ge 0$ and 0 otherwise.

Energy:

$$E = \sum_{n=-\infty}^{\infty}|x[n]|^2 = \sum_{n=0}^{\infty}1 = \infty$$

Power:

$$\begin{aligned} P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}|x[n]|^2 = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=0}^{N}1 \\ &= \lim_{N\to\infty}\frac{N+1}{2N+1} = \frac{1}{2} \end{aligned}$$

Answer: $E = \infty$, $P = 1/2$; $x[n]$ is a power signal.