A linear operator must meet the following conditions:
\[ S\{\alpha x[n]\} = \alpha S\{x[n]\} \]
\[ S\{x[n] + y[n] \} = S\{x[n] \} + S\{y[n] \} \]
Mes: julio 2018
\[ E_x = \sum_{n=-\infty}^{+\infty} \left| x\left[ n \right] \right|^2 \]
\[ P_x = \lim_{N\to\infty} \frac{1}{2N + 1} \sum_{n = -N}^{N} \left| x[n] \right|^2\]
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.