Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

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

schematic

simulate this circuit – Schematic created using CircuitLab

share|improve this question

3 Answers 3

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

share|improve this answer

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

share|improve this answer

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}\$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.