Transfer function

RLC circuits are resonant circuits, as the energy in the system “resonates” between the inductor and capacitor. Shows the math of RLC resonators and visualizes the poles in the Laplace domain. Examines and visualizes the step and frequency response.

RLC circuits are resonant circuits, as the energy in the system “resonates” between the inductor and capacitor. There is an exact analogy between the charge in an RLC circuit and the distance in a mechanical harmonic oscillator (mass attached to spring). We will examine the properties of a resonator consisting of series circuit of an inductor (L), capacitor (C) and resistor (R), where the output is taken across the resistor. The components are the same as in the RLC filter discussed earlier, except that the output is taken over the resistor.\(\)

\(u(t)\)Instead of \(\Delta v(t)\), we use the European symbol for voltage difference: \(u\). The letter ‘u’ stands for “Potentialunterschied”.

Prerequisite reading includes Impedance and Transfer Functions.

own work
LCR schematic

Transfer Function

In the RLC circuit, the current is the input voltage divided by the sum of the impedance of the inductor \(Z_l=j\omega L\), capacitor \(Z_c=\frac{1}{j\omega C}\) and the resistor \(Z_r=R\). The output is the voltage over the capacitor and equals the current through the system multiplied with the capacitor impedance.

$$ \begin{align} H(s) &= \frac{Z_r}{Z_l+Z_c+Z_r}\nonumber \\ &=\frac{R}{sL+\frac{1}{sC}+R}\nonumber \\ &= \frac{R}{L}\frac{s}{s^2+s\frac{R}{L}+\frac{1}{LC}}\label{eq:voltagedivider} \end{align} $$

The zeros of \(H(s)\) are the values of \(s\) such that H(s)=0. There is one zero at \(s=0\). The poles of \(H(s)\) are those values of \(s\) such that \(H(s)=\infty\) . By the quadratic formula, we find the system’s poles:

$$ \begin{align} H(s) &= \frac{R}{L}\frac{(s-z)}{(s-p_1)(s-p-2)} \label{eq:transferpolynomial} \\ \rm{where}\quad z &= 0, \nonumber \\ p_{1,2} &= -\frac{R}{2L}\pm\sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}} \nonumber \end{align} $$

The poles in \(\eqref{eq:transferpolynomial}\) may be real or complex conjugates. Highlight this by parameterizing the transfer function in terms of the damping ratio \(\zeta\), and natural frequency \(\omega_n\). The parameter choices will become evident as we examine complex conjugate poles. [MIT-me]

$$ \begin{align} H(s) &= \frac{R}{L}\frac{s}{s^2+2\zeta\omega_ns+{\omega_n}^2} \label{eq:transferpoles} \\ \rm{where}\quad w_n &= \frac{1}{\sqrt{LC}}, \nonumber \\ \zeta &= \frac{R}{2}\sqrt{\frac{C}{L}} \nonumber \end{align} $$

For resonators (narrow band-pass filters), we commonly use a \(Q\) factor instead of \(\zeta\). The Q factor expresses how under-damped a resonator is, and is defined as the frequency \(\omega\) multiplied with the quotient of the maximum energy stored and the power loss. The Q factor depends on frequency but it is most often quoted for the resonant frequency \(\omega_n\). The maximum energy stored can be calculated from the maximum energy in the inductor. Power is only dissipated in the resistor. For this series RLC circuit, the Q-factor is

$$ \begin{align} Q(\omega_n) &= \omega_n\frac{L\, i_{rms}^2}{R\,i_{rms}^2} \nonumber \\ &= \omega_n\frac{L}{R}=\frac{1}{\sqrt{LC}}\frac{L}{R} \nonumber \\ &= \frac{1}{R}\sqrt{\frac{L}{C}} \end{align} $$

Continue reading about Frequency response.

4 Replies to “Transfer function”

  1. Hi Coert,

    Excellent walk-through!

    Can I ask why your reason for using abs(H(t)) in your bode plots rather than abs(H(w))?

    Another question: I am putting together teaching material for internal education use at my company, is it okay if I use your pictures?

    Best regards,

  2. I want to ask
    What if it’s the inductor that is at vø, I mean if the inductor is where the resistor is

  3. Swapping the components will change the equation (1) for the voltage divider, but the methodology will remain the same.

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