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Sorry if the question is pretty trivial but can anyone tell me what rule or law was used to go from $$e_2=C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4-R_{7}R_2^{-1}e_2$$ $$to$$ $$e_2=\dfrac{R_2*C_4^{-1}}{R_2+R_7}x_1+\dfrac{R_2*C_{10}^{-1}}{R_2+R_7}x_2-\dfrac{R_2*R_{7}*I_3^{-1}}{R_2+R_7}x_4$$

CptPackage
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  • Just as an aside, here on MSE is important that when you feel you had an answer that solved you're problem, you mark it with the green check (accept it)! That gives points to the user who gave the answer as well as letting other MSE user know that this post has been answered correctly – Davide Morgante Jul 19 '18 at 16:27

2 Answers2

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$e_2=C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4-R_{7}R_2^{-1}e_2$

$e_2+R_{7}R_2^{-1}e_2 = C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4$

$e_2(1 + R_{7}R_2^{-1})= C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4$

$e_2 = \frac{C_4^{-1}x_1}{1 + R_{7}R_2^{-1}}+\frac{C_{10}^{-1}x_2}{1 + R_{7}R_2^{-1}}-\frac{R_{7}I_3^{-1}x_4}{1 + R_{7}R_2^{-1}}$

Multiply each numerator and denominator by $R_2$ to get.........

$e_2=\dfrac{R_2*C_4^{-1}}{R_2+R_7}x_1+\dfrac{R_2*C_{10}^{-1}}{R_2+R_7}x_2-\dfrac{R_2*R_{7}*I_3^{-1}}{R_2+R_7}x_4$

Phil H
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$$e_2 = C_4^{-1}x_1+C_10^{-1}x_2-R_7I_3^{-1}x_4\color{red}{-R_7R_2^{-1}e_2} \\ e_2\color{red}{+R_7R_2^{-1}e_2} = C_4^{-1}x_1+C_{10}^{-1}x_2-R_7I_3^{-1} \\ \color{blue}{(1+R_7R_2^{-1})}e_2 = C_4^{-1}x_1+C_{10}^{-1}x_2-R_7I_3^{-1}x_4 \\ e_2 = \frac{C_4^{-1}}{\color{blue}{(1+R_7R_2^{-1})}}\color{red}{\frac{R_2}{R_2}}x_1 + \frac{C_{10}^{-1}}{\color{blue}{(1+R_7R_2^{-1})}}\color{red}{\frac{R_2}{R_2}}x_2-\frac{R_7I_3^{-1}}{\color{blue}{(1+R_7R_2^{-1})}}\color{red}{\frac{R_2}{R_2}}x_4 \\ e_2=\frac{C_4^{-1}R_2}{R_2+R_7}x_1+\frac{C_{10}^{-1}R_2}{R_2+R_7}x_2-\frac{R_3I_3^{-1}R_2}{R_2+R_7}x_4$$

Davide Morgante
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  • Grazie mille, ma la cosa è xk abbiamo multiplicato per $R_2$ per il numeratore e il dinomeratore? – CptPackage Jul 19 '18 at 16:22
  • Siccome al denominatore di ogni addendo compariva il termine $$R_7R_2^{-1}=\frac{R_7}{R_2}$$ In pratica non abbiamo fatto altro che semplificarci i conti. Avremmo potuto benissimo ottimizzare la cosa utilizzando MCM. Ti faccio il conto per un solo addendo, gli altri sono identici $$\frac{C_4^{-1}}{1+\frac{R_7}{R_2}} = \frac{C_4^{-1}}{\frac{R_2+R_7}{R_2}}$$ dopodiché, utilizzando il fatto che $$\frac{a}{\frac{b}{c}} = \frac{ac}{b}$$ possiamo arrivare alla forma finale $$\frac{C_4^{-1}}{\frac{R_2+R_7}{R_2}} = \frac{C_4^{-1}R_2}{R_2+R_7}$$ – Davide Morgante Jul 19 '18 at 16:26
  • Grazie mille, un saluto da Tor Vergata. Ho segnato la tua risposta come la risposta giusta. – CptPackage Jul 19 '18 at 16:38
  • Ahah! Grazie mille, un saluto dalla Sapienza :P – Davide Morgante Jul 19 '18 at 16:38