## Linear operator

A linear operator must meet the following conditions:
$S\{\alpha x[n]\} = \alpha S\{x[n]\}$
$S\{x[n] + y[n] \} = S\{x[n] \} + S\{y[n] \}$

## Energy and power of a signal

$E_x = \sum_{n=-\infty}^{+\infty} \left| x\left[ n \right] \right|^2$
$P_x = \lim_{N\to\infty} \frac{1}{2N + 1} \sum_{n = -N}^{N} \left| x[n] \right|^2$

## Not all sinusoids are periodic in discrete time

For a signal to be periodic, it must fulfill the following condition:

$x[n] = x[n+M]$
where $$M \in \mathbb{Z}$$.

If $$x[n] = e^{j\left(\omega n + \phi\right)}$$, then:

$e^{j\left(\omega n + \phi\right)} = e^{j\left(\omega \left(n+M\right) + \phi\right)}$

$e^{j\omega n } e^{j\phi} = e^{j\omega n }e^{j\omega M} e^{j\phi}$

$1 =e^{j\omega M}$

For $$e^{j \omega M}$$ to be 1, the angle must be multiple of $$2\pi$$. Then:

$\omega M = 2\pi N$

where $$N \in \mathbb{Z}$$ and $$2 \pi N$$ represents any integer multiple of $$2\pi$$.
Then, the frequency must meet:
$\omega = \frac{2\pi N}{M}$

This is, $$\omega$$ must be a rational multiple of $$2\pi$$.

What does this mean?

Basically, that a periodic signal can’t take any arbitrary frequency in discrete time.