La funzione di trasferimento è il rapporto tra uscita e ingresso di un sistema
LTI sul dominio delle frequenze.
Possiamo applicarla al caso dei fasori di corrente visti nel circuito RLC in
parallelo nella scorsa lezione.
H ~ ( ω ) = uscita ingresso = I ~ R 1 I ~ s \tilde{H}(\omega) = \frac{\text{uscita}}{\text{ingresso}} = \frac{\tilde{I}_{R_1}}{\tilde{I}_s} H ~ ( ω ) = ingresso uscita = I ~ s I ~ R 1 Esempio : il circuito visto la scorsa lezione:
H ~ ( ω ) = v ~ s R v ~ s Z tot = 1 1 R + j ω C + 1 j ω L R = 1 R ( 1 R + j ω C − j 1 ω L ) = 1 R ( 1 R + j ( ω C − 1 ω L ) ) = 1 1 + j R ( ω C − 1 ω L ) = 1 1 + j R ( ω C − 1 ω L ) \begin{array}{ll}
\tilde{H}(\omega) &= \frac{\frac{\tilde{v}_s}{R}}{\frac{\tilde{v}_s}{Z_{\text{tot}}}} \\[7pt]
&= \frac{\frac{1}{\frac{1}{R} + j \omega C + \frac{1}{j \omega L}}}{R} \\[7pt]
&= \frac{1}{R (\frac{1}{R} + j \omega C - j \frac{1}{\omega L})} \\[7pt]
&= \frac{1}{R (\frac{1}{R} + j (\omega C - \frac{1}{\omega L}))} \\[7pt]
&= \frac{1}{1 + j R (\omega C - \frac{1}{\omega L})} \\[7pt]
&= \frac{1}{1 + j R (\omega C - \frac{1}{\omega L})}
\end{array} H ~ ( ω ) = Z tot v ~ s R v ~ s = R R 1 + j ω C + j ω L 1 1 = R ( R 1 + j ω C − j ω L 1 ) 1 = R ( R 1 + j ( ω C − ω L 1 )) 1 = 1 + j R ( ω C − ω L 1 ) 1 = 1 + j R ( ω C − ω L 1 ) 1 A questo punto sostituiamo ω 0 = 1 L C \omega_0 = \frac{1}{\sqrt{L C}} ω 0 = L C 1 Q = R C L Q = R \sqrt{\frac{C}{L}} Q = R L C R C = R C L L C = Q ω 0 R C = R \sqrt{\frac{C}{L}} \sqrt{L C} = \frac{Q}{\omega_0} R C = R L C L C = ω 0 Q
H ~ ( ω ) = 1 1 + j R C ( ω − 1 ω L C ) = 1 1 + j R C ( ω − ω 0 2 ω ) = 1 1 + j Q ( ω ω 0 − ω 0 ω ) 1 − j Q ( ω ω 0 − ω 0 ω ) 1 − j Q ( ω ω 0 − ω 0 ω ) = 1 − j Q ( ω ω 0 − ω 0 ω ) 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 \newcommand\Xw{\left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right)}
\begin{array}{ll}
\tilde{H}(\omega) &= \frac{1}{1 + j R C (\omega - \frac{1}{\omega L C})} \\[10pt]
&= \frac{1}{1 + j R C (\omega - \frac{\omega_0^2}{\omega})} \\[10pt]
&= \frac{1}{1 + j Q \Xw} \frac{1 - j Q \Xw}{1 - j Q \Xw} \\[10pt]
&= \frac{1 - j Q \Xw}{1 + Q^2 \Xw^2}
\end{array} H ~ ( ω ) = 1 + j R C ( ω − ω L C 1 ) 1 = 1 + j R C ( ω − ω ω 0 2 ) 1 = 1 + j Q ( ω 0 ω − ω ω 0 ) 1 1 − j Q ( ω 0 ω − ω ω 0 ) 1 − j Q ( ω 0 ω − ω ω 0 ) = 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 1 − j Q ( ω 0 ω − ω ω 0 ) Il modulo della funzione definisce il guadagno del circuito:
G ( ω ) = ∣ H ~ ( ω ) ∣ = 1 − j Q ( ω ω 0 − ω 0 ω ) 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 1 + j Q ( ω ω 0 − ω 0 ω ) 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 = ( 1 − j Q ( ω ω 0 − ω 0 ω ) ) ( 1 + j Q ( ω ω 0 − ω 0 ω ) ) 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 = 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 = 1 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 \newcommand\Xw{\left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right)}
\begin{array}{ll}
G(\omega) = |\tilde{H}(\omega)| &= \sqrt{\frac{1 - j Q \Xw}{1 + Q^2 \Xw^2} \frac{1 + j Q \Xw}{1 + Q^2 \Xw^2}} \\[15pt]
&= \frac{\sqrt{\left(1 - j Q \Xw\right) \left(1 + j Q \Xw\right)}}{1 + Q^2 \Xw^2} \\[15pt]
&= \frac{\sqrt{1 + Q^2 \Xw^2}}{1 + Q^2 \Xw^2} \\[15pt]
&= \frac{1}{\sqrt{1 + Q^2 \Xw^2}}
\end{array} G ( ω ) = ∣ H ~ ( ω ) ∣ = 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 1 − j Q ( ω 0 ω − ω ω 0 ) 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 1 + j Q ( ω 0 ω − ω ω 0 ) = 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 ( 1 − j Q ( ω 0 ω − ω ω 0 ) ) ( 1 + j Q ( ω 0 ω − ω ω 0 ) ) = 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 = 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 1 Il cambiamento di fase tra l’ingresso e l’uscita è dato da:
θ ( ω ) = arg ( H ~ ( ω ) ) = arctan ( ( 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 ) ( − Q ( ω ω 0 − ω 0 ω ) ) 1 + Q 2 ( ω ω 0 − ω 0 ω ) 2 ) = arctan ( − Q ( ω ω 0 − ω 0 ω ) 1 ) = − arctan ( Q ( ω ω 0 − ω 0 ω ) ) \newcommand\Xw{\left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right)}
\begin{array}{ll}
\theta(\omega) = \arg{\tilde{H}(\omega)} &= \arctan\left(\frac{\left(1 + Q^2\ \Xw^2\right) \left(- Q \Xw\right)}{1 + Q^2\ \Xw^2}\right) \\[15pt]
&= \arctan\left(\frac{-Q\ \Xw}{1}\right) \\[10pt]
&= -\arctan\left(Q\ \Xw\right)
\end{array} θ ( ω ) = arg ( H ~ ( ω ) ) = arctan 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 ( 1 + Q 2 ( ω 0 ω − ω ω 0 ) 2 ) ( − Q ( ω 0 ω − ω ω 0 ) ) = arctan ( 1 − Q ( ω 0 ω − ω ω 0 ) ) = − arctan ( Q ( ω 0 ω − ω ω 0 ) ) X RMS = 1 T ∫ 0 T A 2 cos ( ω t + θ ) 2 d t = ω A 2 2 π ∫ 0 2 π ω cos ( ω t + θ ) 2 d t = ω A 2 2 π ∫ 0 2 π ω cos ( 2 ω t + 2 θ ) + 1 2 d t = ω A 2 4 π 2 π ω = A 2 2 = A 2 \begin{array}{ll}
X_{\text{RMS}} &= \sqrt{\frac{1}{T} \int_0^T A^2\ \fCos{\omega t + \theta}^2\ dt} \\[10pt]
&= \sqrt{\frac{\omega A^2}{2 \pi} \int_0^{\frac{2 \pi}{\omega}} \fCos{\omega t + \theta}^2\ dt} \\[10pt]
&= \sqrt{\frac{\omega A^2}{2 \pi} \int_0^{\frac{2 \pi}{\omega}} \frac{\fCos{2 \omega t + 2 \theta} + 1}{2} dt} \\[10pt]
&= \sqrt{\frac{\omega A^2}{4 \pi} \frac{2 \pi}{\omega}} \\[10pt]
&= \sqrt{\frac{A^2}{2}} = \frac{A}{\sqrt{2}}
\end{array} X RMS = T 1 ∫ 0 T A 2 cos ( ω t + θ ) 2 d t = 2 π ω A 2 ∫ 0 ω 2 π cos ( ω t + θ ) 2 d t = 2 π ω A 2 ∫ 0 ω 2 π 2 cos ( 2 ω t + 2 θ ) + 1 d t = 4 π ω A 2 ω 2 π = 2 A 2 = 2 A Nei circuiti a corrente alternata, il valore RMS di V V V V V V