Esercizio 5 -

Usando la notazione polare per i numeri complessi:

$\frac{v_2}{v_1} = \frac{565 e^{j45^\circ}}{20} = 28.25 e^{j45^\circ} = 28.25(\cos 45^\circ + j \sin 45^\circ) = (20 + j20)$
$\overline{Z} = \frac{1}{j\omega C} = \frac{1}{10^5 \cdot 20 \cdot 10^{-12}} = \frac{10^7}{20j} = -j 5 \cdot 10^5\ \Omega$
$\overline{Z}_1 = \frac{\overline{Z}}{1-\tfrac{v_2}{v_1}} \;\;\Rightarrow\;\; \overline{Y}_1 = \frac{1-\tfrac{v_2}{v_1}}{\overline{Z}} = \frac{1+20+j20}{-j5 \cdot 10^5} = \frac{21+j20}{-j5 \cdot 10^5} = (4+j4.2)\cdot 10^{-5}\ \text{S}$
$G_1 = -40 \cdot 10^{-6}\ \text{S}, \qquad B_1 = 42 \cdot 10^{-6}\ \text{S}$
$\overline{Y}_2 = \frac{1-\tfrac{v_1}{v_2}}{\overline{Z}} = \frac{1+\tfrac{1}{20+j20}}{-j5\cdot 10^5} = \frac{21+j20}{-j5\cdot 10^5 (20+j20)} = \frac{21+j20}{100 - j100} \cdot 10^{-5} = \frac{21+j20}{1-j} \cdot 10^{-7} = \frac{(1+j)(21+j20)}{2} \cdot 10^{-7} = \frac{21+j20+j21-20}{2} \cdot 10^{-7} = \frac{1+j41}{2} \cdot 10^{-7} = (0.5 + j20.5)\cdot 10^{-7}$

$G_2 = 0.05 \cdot 10^{-6}\ \text{S}, \qquad B_2 = 2.05 \cdot 10^{-6}\ \text{S}$

Esercizio 5

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