CT unit step; periodicity of e^(jωt)

CT unit step

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

It switches from 0 to 1 at $t = 0$; $u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau$.

Periodicity of $x(t) = e^{j\omega t}$

$x(t)$ is periodic with period $T$ if $x(t+T) = x(t)$ for all $t$:

$$e^{j\omega(t+T)} = e^{j\omega t}e^{j\omega T} = e^{j\omega t} \;\Rightarrow\; e^{j\omega T} = 1$$ $$\cos\omega T + j\sin\omega T = 1 \;\Rightarrow\; \omega T = 2\pi m, \quad m = \pm1, \pm2, \ldots$$

So the necessary condition is $\omega T = 2\pi m$. For $\omega \ne 0$ the smallest positive $T$ (with $m = 1$) is

$$T_0 = \frac{2\pi}{|\omega|}$$

Since $T$ can be any real number, this is satisfied for every non-zero $\omega$: a CT complex exponential is always periodic (for $\omega = 0$ it is a constant, periodic with any $T$). This differs from DT, where $\omega/2\pi$ must also be rational.