## 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.

## 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}

## Transfer function

Filters can remove low and/or high frequencies from an electronic signal. My article on RC Low-pass Filter introduced a first order low-pass filter. The second order filter introduced here improves the unit step response and the the roll-off slope for the frequency response. As we will learn, even this passive filters may exhibit resonance near the natural frequency.

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

We examine the properties of a filter consisting of a series circuit of an inductor $$L$$, resistor $$R$$ and capacitor $$C$$. The output is taken across the capacitor as shown in the schematic below. We show the transfer function and derive the step and frequency response.

This article describes a low-pass filter, but the same principles apply to high and band pass filters and can even be extended to to resonators. For example, taking the voltage over the inductor results in a high-pass filter, while taking the voltage over the resistor makes a band-pass filter.

Prerequisite reading includes Laplace Transforms, Impedance and Transfer Functions.

## Transfer Function

In the RLC circuit, shown above, the current is the input voltage divided by the sum of the impedance of the inductor $$Z_L$$, the impedance of the resistor $$Z_R=R$$ and that of the capacitor $$Z_C$$. The output is the voltage over the capacitor and equals the current through the system multiplied with the capacitor impedance.

\begin{align} Y(s) &= I(s)Z_C=\frac{U(s)}{Z_L+Z_R+Z_C} Z_C \Rightarrow \nonumber \\ H(s) &= \frac{Y(s)}{U(s)}=\frac{Z_C}{Z_L+Z_R+Z_C} =\frac{\frac{1}{sC}}{sL+R+\frac{1}{sC}} = \frac{1}{s^2LC+sRC+1} \label{eq:voltagedivider} \end{align}

The denominator of $$\eqref{eq:voltagedivider}$$ is a second-order polynomial. The roots to this polynomial are called the system’s poles.

$$\shaded{ H(s) = K\frac{1}{(s-p_1)(s-p_2)} } \label{eq:transferpolynomial}$$
where
\begin{align*} K &= \frac{1}{LC} \\ p_{1,2} &= -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}} \end{align*}

Note that the factor $$K$$ can be expressed as the product $$p_1\cdot p_2$$ by applying the identity $$(a+b)(a-b)=a^2-b^2$$

$$K = p_1\cdot p_2=\frac{1}{LC} \label{eq:p1p2}$$

Here is where is gets interesting:

The poles in $$\eqref{eq:transferpolynomial}$$ may be real or complex conjugates.

To highlight this, we parameterize the poles in terms of the natural frequency $$\omega_n$$ and damping ratio $$\zeta$$.

\begin{align} p_{1,2} &= -\frac{R}{2L} \pm \sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}}\nonumber\\ &= \frac{1}{\sqrt{LC}} \left( -\sqrt{LC}\frac{R}{2L} \pm \sqrt{LC}\sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}} \right) \nonumber \\ &= \underbrace{\frac{1}{\sqrt{LC}}}_{\omega_n} \left( -\underbrace{\frac{R}{2}\sqrt\frac{C}{L}}_{\zeta} \pm \sqrt{\underbrace{\left(\frac{R}{2}\sqrt\frac{C}{L}\right)^2}_{\zeta^2}-1} \right) \\ \end{align}

Assign the natural frequency $$\omega_n=\sqrt{\frac{1}{LC}}$$ and the damping ratio $$\zeta=\frac{R}{2}\sqrt{\frac{C}{L}}$$. These parameter choices will become evident as we examine complex conjugate poles.

$$\shaded{ H(s) = K\frac{1}{(s-p_1)(s-p_2)} } \label{eq:transferpoles}$$
where
\begin{align*} K &= \frac{1}{LC} \\ p_{1,2} &= \omega_n\left(-\zeta \pm \sqrt{\zeta^2-1}\right) \\ \zeta &= \frac{R}{2}\sqrt{\frac{C}{L}} \\ \omega_n &= \sqrt{\frac{1}{LC}} \end{align*}

This dampening coefficient $$\zeta$$ determines the behavior of the system. Given that the value of $$\frac{1}{LC} >0$$ and $$\frac{R}{L}\geq0$$, we can identify three classes of pole locations.

Effect of dampening coefficient $$\zeta$$ on system behavior
ζ Pole location Referred to as
$$\zeta>1$$ different locations on the negative real axis overdamped $$R>2\sqrt\frac{L}{C}$$
$$\zeta=1$$ coincite on the negative real axis critically damped $$R=2\sqrt\frac{L}{C}$$
$$\zeta<1$$ complex conjugate poles in the left half of the s-plane underdamped $$R<2\sqrt\frac{L}{C}$$

In the overdamped case, $$\zeta>1$$, the following relation between $$R$$, $$L$$ and $$C$$ must be true

\begin{align} \zeta&>1\nonumber\\ \implies \frac{R}{2}\sqrt{\frac{C}{L}}&>1 \nonumber \\ \implies R&>2\sqrt\frac{L}{C} \end{align}

Following sections determine the step and frequency response for each of these classes.
• Two different real poles
• Coinciting real poles
• Complex poles

## Transfer function

Shows the math of a first order RC low-pass filter. Visualizes the poles in the Laplace domain. Calculates and visualizes the step and frequency response.

Filters can remove low and/or high frequencies from an electronic signal, to suppress unwanted frequencies such as background noise. This article shows the math and visualizes the system response of such filters.

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

One of the simplest forms of passive filters consists of a resistor and capacitor in series. The output is the voltage over the capacitor $$y(t)$$ as shown in the schematic below.

This type of filter is called an Infinite-Impulse Response (IIR) filter, because if you give it an impulse input, the output takes an infinite time to go down to exactly zero.

Even though this article shows a low pass filter, the same principles apply to a high pass filter where the output is taken over the resistor.

We will derive the transfer function for this filter and determine the step and frequency response functions. Required prior reading includes Laplace Transforms, Impedance and Transfer Functions.

In this article will will use Laplace Transforms. The alternate method of solving the linear differential equation is shown in Appendix B for reference.

## Transfer Function

In the RC circuit, shown above, the current is the input voltage divided by the sum of the impedance of the resistor $$Z_R=R$$ and that of the capacitor $$Z_C$$. The output is the voltage over the capacitor and equals the current through the system multiplied with the capacitor impedance. The transfer function follows as the quotient of the output and input signals.

$$H(s) = \frac{Y(s)}{U(s)} = \frac{Z_C}{Z_C+Z_R} = \frac{1/sC}{1/sC+R} = \frac{1}{sRC+1} \label{eq:voltagedivider}$$

The denominator of $$\eqref{eq:voltagedivider}$$ is a first-order polynomial. The root of this polynomial is called the system’s pole

$$\shaded{ H(s) = K\frac{1}{s-p},\quad K = \frac{1}{RC},\quad p = -\frac{1}{RC} } \label{eq:transferpolynomial}$$

This first order system $$\eqref{eq:transferpolynomial}$$ has no zeros and one stable pole $$p$$ on the left real axis $$p\lt 0$$ as visualized in the $$s$$-plane.

Moving on from this “unit step response of RC low-pass filter”, continue reading about the Unit Step Response.