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How can we find the transfer function $$\frac{E_0(s)}{E_1(s)}$$ of this RC network


simulate this circuit – Schematic created using CircuitLab

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up vote 3 down vote accepted

Replace the capacitor C with an impedance of value \$Z_c = 1/(Cs)\$ and apply voltage division rule.

ie, $$ E_0(s) = E_1(s) \times \frac{Z_c}{R+Z_c}$$

or, $$\frac{E_0(s)}{E_1(s)} = \frac{Z_c}{Z_c+R}$$

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With resistor potential dividers like this: -

enter image description here

It's exactly the same for any impedance. On your circuit, the impedance at the bottom is \$\dfrac{1}{s\cdot C}\$ where s is \$2\cdot\pi\cdot f\$.

Therefore the transfer function is: -

\$\dfrac{\dfrac{1}{s\cdot C}}{{\dfrac{1}{s\cdot C} + R}}\$ and this reduces to: -

\$\dfrac{1}{1+s\cdot C\cdot R}\$

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$$\frac{E0(s)}{E1(s)}=\frac{\frac{1}{s \cdot C}}{R+\frac{1}{s \cdot C}} = \frac{1}{1+s \cdot C \cdot R}$$

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