Derive Transfer Function of Electrical Circuit Using Impedances – Fusion of Engineering, Control, Coding, Machine Learning, and Science

In this tutorial, we explain how to derive a transfer function of electrical circuits using impedances. As an example, we use an RLC (resistor, inductor, and capacitor) circuit. The YouTube tutorial is given below.

Problem Formulation

Consider the circuit shown below consisting of the capacitor with capacitance $C$ , inductor with inductance of $L$ , and the resistor with the resistance of $R$ . Let $v_{i}(t)$ be the input voltage and $v_{o}(t)$ be the output voltage. Derive the transfer function of this circuit from the input voltage $v_{i}(t)$ to the output voltage $v_{o}(t)$ .

Derivation of the transfer function using impedance approach

To solve this problem, we use the impedance approach. The idea is to convert this circuit in the complex s-domain and use impedances. The impedances are explained in our previous tutorial given here. The circuit in the s-domain is shown below.

In the figure above, $Z_{1}(s)$ is the impedance of the inductor, $Z_{2}(s)$ is the impedance of the resistor $R$ , and $Z_{3}(s)$ is the impedance of the capacitor. As shown in our previous tutorial, we have

(1) $\begin{align*}Z_{1}(s)= Ls \\Z_{2}(s)=R \\Z_{3}(s)=\frac{1}{Cs}\end{align*}$

In the figure above, $V_{i}(s)$ is the Laplace transform of the input voltage $v_{i}(t)$ and $V_{o}(s)$ is the Laplace transform of the output voltage $v_{o}(t)$ . Finally, $V_{A}(s)$ is the Laplace transform of the voltage potential at the point $A$ . The transfer function is defined by

(2) $\begin{align*}W(s)=\frac{V_{i}(s)}{V_{o}(s)}\end{align*}$

To derive this transfer function, let us first simplify the notation by dropping out the dependencies on the complex s-variable. For example, instead of $Z_{1}(s)$ , we simply write $Z_{1}$ . Keep in mind that we use the standard notation of using capital letters for Laplace transformed variables and lower-case letters for the time domain signals. From the figure given above, we have:

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

From the last equation, we have

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

Finally, we get

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

On the other hand, we have

(6) $\begin{align*}\frac{V_{o}}{Z_{3}}=\frac{V_{A}-V_{o}}{Z_{2}} \\V_{o}Z_{2}=(V_{A}-V_{o})Z_{3} \\V_{0}(Z_{2}+Z_{3})=V_{A}Z_{3} \\V_{A}=\frac{Z_{2}+Z_{3}}{Z_{3}}V_{o}\end{align*}$

By substituting the last equation in (5), and after several manipulations, we obtain the transfer function that depends on impedances

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

By substituting impedances for the resistor, capacitor, and inductor in the equation (7), we obtain the transfer function expression

(8) $\begin{align*}\frac{V_{o}}{V_{i}}=W(s)=\frac{\frac{1}{Cs}}{Ls+R+\frac{1}{Cs}}=\frac{1}{LCs^{2}+RCs+1}\end{align*}$

Problem Formulation

Derivation of the transfer function using impedance approach

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