Periodicity of 3e^(j3π(n+1/2)/5)

Necessary condition

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.

Is $x[n] = 3e^{j3\pi(n + 1/2)/5}$ periodic?

Separate the constant phase:

$$x[n] = 3e^{j3\pi/10}\,e^{j\frac{3\pi}{5}n}$$

$3e^{j3\pi/10}$ is a constant complex amplitude, so periodicity depends on $\omega_0 = \frac{3\pi}{5}$:

$$\frac{\omega_0}{2\pi} = \frac{3\pi/5}{2\pi} = \frac{3}{10} \quad\text{(rational)}$$

So the signal is periodic. Fundamental period:

$$N = \frac{2\pi m}{\omega_0} = \frac{10m}{3} \;\Rightarrow\; N = 10 \text{ (with } m = 3\text{)}$$

Check: $x[n+10] = x[n]\,e^{j\frac{3\pi}{5}\cdot 10} = x[n]e^{j6\pi} = x[n]$.

Answer: periodic with fundamental period $N = 10$ (fundamental frequency $2\pi/10 = \pi/5$ rad/sample).