Potenza di un segnale dati

0.350000

Si è affermato che la potenza di un segnale

s$$\left(\vphantom{ t}\right.$$t$$\left.\vphantom{ t}\right)$$ = $$\sum_{n}^{}$$a_ng$$\left(\vphantom{ t-nT}\right.$$t - nT$$\left.\vphantom{ t-nT}\right)$$

in cui g$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ è una caratteristica di Nyquist a coseno rialzato con roll-off $\gamma$, e gli a_i sono una sequenza di v.a. discrete, statisticamente indipendenti, ed uniformemente distribuite su L livelli in una dinamica - ${\frac{\Delta }{2}}$ $\leq$ a_i $\leq$ ${\frac{\Delta }{2}}$, ha valore

$$\mathcal {P}$$_s = $${\frac{\Delta ^{2}}{12}}$$$${\frac{L+1}{L-1}}$$$$\left(\vphantom{ 1-\frac{\gamma }{4}}\right.$$1 - $${\frac{\gamma }{4}}$$ $$\left.\vphantom{ 1-\frac{\gamma }{4}}\right)$$

Nella precedente appendice, si è mostrato che per lo stesso segnale risulta $\mathcal {P}$_s$\left(\vphantom{ f}\right.$f$\left.\vphantom{ f}\right)$ = $\sigma_{A}^{2}$${\frac{\left\vert G\left( f\right) \right\vert ^{2}}{T}}$, e dunque

$$\mathcal {P}$$_s = $$\int$$$$\mathcal {P}$$_s$$\left(\vphantom{ f}\right.$$f$$\left.\vphantom{ f}\right)$$df = $$\int$$$$\sigma_{A}^{2}$$$${\frac{\left\vert G\left( f\right) \right\vert ^{2}}{T}}$$df

Svolgendo i relativi calcoli, si può mostrare che $\int$$\left\vert\vphantom{ G\left( f\right) }\right.$G$\left(\vphantom{ f}\right.$f$\left.\vphantom{ f}\right)$ $\left.\vphantom{ G\left( f\right) }\right\vert^{2}_{}$df = T$\left(\vphantom{ 1-\frac{\gamma }{4}}\right.$1 - ${\frac{\gamma }{4}}$ $\left.\vphantom{ 1-\frac{\gamma }{4}}\right)$, e quindi $\mathcal {P}$_s = $\sigma_{A}^{2}$$\left(\vphantom{ 1-\frac{\gamma }{4}}\right.$1 - ${\frac{\gamma }{4}}$ $\left.\vphantom{ 1-\frac{\gamma }{4}}\right)$; resta pertanto da calcolare $\sigma_{A}^{2}$:

$\sigma_{A}^{2}$ = (a_i a media nulla) = E_X$\left\{\vphantom{ a_{i}^{2}}\right.$a_i²$\left.\vphantom{ a_{i}^{2}}\right\}$ = $\sum_{i=0}^{L-1}$p_ai ^. a_i² = (a_i equiprobabili) =

= ${\frac{1}{L}}$$\sum_{i=0}^{L-1}$$\left(\vphantom{ i\frac{\Delta }{L-1}-\frac{\Delta }{2}}\right.$i${\frac{\Delta }{L-1}}$ - ${\frac{\Delta }{2}}$ $\left.\vphantom{ i\frac{\Delta }{L-1}-\frac{\Delta }{2}}\right)^{2}_{}$ = ${\frac{\Delta ^{2}}{L}}$$\sum_{i=0}^{L-1}$$\left(\vphantom{ \frac{i^{2}}{\left( L-1\right) ^{2}}+\frac{1}{4}-\frac{i}{L-1}}\right.$${\frac{i^{2}}{\left( L-1\right) ^{2}}}$ + ${\frac{1}{4}}$ - ${\frac{i}{L-1}}$ $\left.\vphantom{ \frac{i^{2}}{\left( L-1\right) ^{2}}+\frac{1}{4}-\frac{i}{L-1}}\right)$ =

= ${\frac{\Delta ^{2}}{L}}$$\left(\vphantom{ \frac{L}{4}-\frac{1}{L-1}\sum _{i=0}^{L-1}i+\frac{1}{\left( L-1\right) ^{2}}\sum _{i=0}^{L-1}\left( i\right) ^{2}}\right.$${\frac{L}{4}}$ - ${\frac{1}{L-1}}$$\sum_{i=0}^{L-1}$i + ${\frac{1}{\left( L-1\right) ^{2}}}$$\sum_{i=0}^{L-1}$$\left(\vphantom{ i}\right.$i$\left.\vphantom{ i}\right)^{2}_{}$ $\left.\vphantom{ \frac{L}{4}-\frac{1}{L-1}\sum _{i=0}^{L-1}i+\frac{1}{\left( L-1\right) ^{2}}\sum _{i=0}^{L-1}\left( i\right) ^{2}}\right)$ =

(facendo uso delle relazioni $\sum_{n=1}^{N}$n = ${\frac{N\left( N+1\right) }{2}}$ e $\sum_{n=1}^{N}$n² = ${\frac{N\left( N+1\right) \left( 2N+1\right) }{6}}$ )

= ${\frac{\Delta ^{2}}{L}}$$\left(\vphantom{ \frac{L}{4}-\frac{1}{L-1}\frac{L\left( L-1\right) }{2}+\frac{... ... ^{2}}\frac{\left( L-1\right) L\left( 2\left( L-1\right) +1\right) }{6}}\right.$${\frac{L}{4}}$ - ${\frac{1}{L-1}}$${\frac{L\left( L-1\right) }{2}}$ + ${\frac{1}{\left( L-1\right) ^{2}}}$${\frac{\left( L-1\right) L\left( 2\left( L-1\right) +1\right) }{6}}$ $\left.\vphantom{ \frac{L}{4}-\frac{1}{L-1}\frac{L\left( L-1\right) }{2}+\frac{... ... ^{2}}\frac{\left( L-1\right) L\left( 2\left( L-1\right) +1\right) }{6}}\right)$ = $\Delta^{2}_{}$$\left(\vphantom{ \frac{1}{4}-\frac{1}{2}+\frac{2L-2+1}{6\left( L-1\right) }}\right.$${\frac{1}{4}}$ - ${\frac{1}{2}}$ + ${\frac{2L-2+1}{6\left( L-1\right) }}$ $\left.\vphantom{ \frac{1}{4}-\frac{1}{2}+\frac{2L-2+1}{6\left( L-1\right) }}\right)$ =

= $\Delta^{2}_{}$${\frac{6L-6-12L+12+8L-8+4}{24\left( L-1\right) }}$ = $\Delta^{2}_{}$${\frac{2L+2}{24\left( L-1\right) }}$ = ${\frac{\Delta ^{2}}{12}}$${\frac{L+1}{L-1}}$.