Transfer Function of the RC circuit – Fusion of Engineering, Control, Coding, Machine Learning, and Science

In this electrical engineering tutorial we explain how to derive a transfer function of an RC (resistor and capacitor) circuit. The YouTube video is given below.

Problem Formulation

Consider the circuit shown below. The circuit consists of the resistor with the resistance of $R$ , voltage source, and the capacitor with the capacitance of $C$ . For the input $v_{i}(t)$ and output $v_{o}(t)$ voltages shown in the figure below, derive the transfer function of the RC circuit.

Derivation of the transfer function of the RC circuit

To derive the transfer function, we use the impedance approach. We transform the circuit shown above in the complex $s$ domain. The resistor and capacitor are replaced by the corresponding impedances, and the voltages are replaced by the Laplace transforms of voltages. The transformed circuit is shown below.

In the figure above, the impedance $Z_{1}(s)$ is the impedance of the resistor. This impedance is given by

(1) $\begin{align*}Z_{1}(s)=R\end{align*}$

The impedance $Z_{2}(s)$ is the impedance of the capacitor. This impedance is given by

(2) $\begin{align*}Z_{2}(s)=\frac{1}{Cs}\end{align*}$

The Laplace transform of $v_{i}(t)$ is denoted by $V_{i}(s)$ . The Laplace transform of $v_{o}(t)$ is denoted by $V_{o}(s)$ .

By using the Ohm’s law in the complex Laplace domain, we obtain

(3) $\begin{align*}\frac{V_{i}-V_{o}}{Z_{1}}=\frac{V_{o}}{Z_{2}}\end{align*}$

where for notation simplicity, we drop out the dependence of variables on $s$ . From the last equation, we have

(4) $\begin{align*}(V_{i}-V_{o})Z_{2}=V_{o}Z_{1} \\V_{i}Z_{2}=V_{o}(Z_{1}+Z_{2})\end{align*}$

From the last equation, we obtain

(5) $\begin{align*}\frac{V_{o}}{V_{i}}=\frac{Z_{2}}{Z_{1}+Z_{2}}\end{align*}$

By substituting the impedances in the last equation, we obtain

(6) $\begin{align*}\frac{V_{o}}{V_{i}}=\frac{\frac{1}{Cs}}{R+\frac{1}{Cs}}=\frac{1}{1+RCs}\end{align*}$

The last equation is the transfer function of the RC circuit. Obviously, this is a low-pass filter. The break frequency or the corner frequency is given at

(7) $\begin{align*}\omega & = \frac{1}{RC} \\f&= \frac{1}{2\pi \cdot RC}\end{align*}$

Problem Formulation

Derivation of the transfer function of the RC circuit

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