# 4.6 Second-order filters

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Summarizes second order lowpass, bandpass, and highpass fitlers.

## Second-order lowpass filters

The second-order lowpass filter has system function

$H\left(s\right)=\frac{{\Omega }_{n}^{2}}{{s}^{2}+2\zeta {\Omega }_{n}s+{\Omega }_{n}^{2}}$

where ${\Omega }_{n}$ is the undamped natural frequency , and $\zeta$ is the damping ratio . The undamped natural frequency and damping ratio are properties of the physical devices used to implement the second-order filter (capacitors, inductors, and resistors, in the case of an electrical circuit). It so happens that the damping ratio satisfies $\zeta \ge 0$ . Using the formula for the roots of a quadratic polynomial, the two poles of $H\left(s\right)$ are easily found to be

$\begin{array}{cc}\hfill {s}_{1}& =-\zeta {\Omega }_{n}+{\Omega }_{n}\sqrt{{\zeta }^{2}-1}\hfill \\ \hfill {s}_{2}& =-\zeta {\Omega }_{n}-{\Omega }_{n}\sqrt{{\zeta }^{2}-1}\hfill \end{array}$

There are three possible sets of poles that are categorized as follows:

1. Overdamped: the poles are real and distinct. This occurs if $\zeta >1$ . In this case the impulse response is given by:
$h\left(t\right)=\frac{{\Omega }_{n}}{2\sqrt{{\zeta }^{2}-1}}{e}^{{s}_{1}t}u\left(t\right)-\frac{{\Omega }_{n}}{2\sqrt{{\zeta }^{2}-1}}{e}^{{s}_{2}t}u\left(t\right)$
2. Critically damped: corresponds to $\zeta =1$ . The two poles are repeated with,
${s}_{1}={s}_{2}=-\zeta {\Omega }_{n}=-{\Omega }_{n}$
and the impulse response is given by
$h\left(t\right)={\Omega }_{n}^{2}t{e}^{-\zeta {\Omega }_{n}t}u\left(t\right)$
3. Underdamped: corresponds to $0\le \zeta <1$ , giving a pair of complex conjugate poles. In this case, the impulse response is given by
$h\left(t\right)=\frac{{\Omega }_{n}}{\sqrt{1-{\zeta }^{2}}}{e}^{-\zeta {\Omega }_{n}t}sin\left(\sqrt{1-{\zeta }^{2}},{\Omega }_{n},t\right)u\left(t\right)$
Note that in the underdamped case, the magnitude of the poles is $|{s}_{1}|=|{s}_{2}|={\Omega }_{n}$ , and when $\zeta =0$ , the two poles are on the imaginary axis which corresponds to an impulse response that is a pure sinusoid and the system is unstable.

A root locus diagram shows the paths that the poles of $H\left(s\right)$ would take as the damping ratio is decreased from some number greater than 1 down to 0, and is shown in [link] . Root locus of second-order lowpass filter having Ω n = 10 .

The frequency response of the second-order lowpass filter can be found using the substitution ${H\left(j\Omega \right)=H\left(s\right)|}_{s=j\Omega }$ , giving

$H\left(j,\Omega \right)=\frac{{\Omega }_{n}^{2}}{{\Omega }_{n}^{2}-{\Omega }^{2}+j2\zeta {\Omega }_{n}\Omega }$

The frequency response magnitude is shown in [link] for ${\Omega }_{n}=10$ and several values of $\zeta$ . Note that for the underdamped case, there is a resonance or peak that would be expected to maximize at the undamped natural frequency ${\Omega }_{n}$ , when $\zeta =0$ . Frequency response of second-order lowpass filter with Ω n = 10 and several values of ζ .

We also observe that for the critically damped case, since

$H\left(s\right)=\frac{{\Omega }_{n}^{2}}{{\left(s,+,{\Omega }_{n}\right)}^{2}}$

setting $s=j\Omega$ gives $\left|H,\left({\Omega }_{n}\right)\right|=\frac{1}{2}$ . Moreover, it is clear that for $\Omega <{\Omega }_{n}$ , $\left|H,\left({\Omega }_{n}\right)\right|>\frac{1}{2}$ and for $\Omega >{\Omega }_{n}$ , $\left|H,\left({\Omega }_{n}\right)\right|<\frac{1}{2}$ . These ideas will be used later when we discuss Bode plots.

## Second-order filter implementation

Second order filters can be implemented using either passive or active circuit elements. We will consider the series RLC circuit shown in [link] . A lowpass filter results by taking the filter output to be the voltage across the capacitor. Using voltage division, the system function is easily found to be

$\begin{array}{cc}\hfill {H}_{lp}\left(s\right)& =\frac{{V}_{lp}\left(s\right)}{{V}_{i}\left(s\right)}\hfill \\ & =\frac{\frac{1}{LC}}{{s}^{2}+\frac{R}{L}s+\frac{1}{LC}}\hfill \end{array}$

Comparing [link] with [link] we find that the undamped natural frequency is ${\Omega }_{n}=\frac{1}{\sqrt{LC}}$ , while $2\zeta {\Omega }_{n}=\frac{R}{L}$ , giving the attenuation coefficient:

$\alpha \equiv \zeta {\Omega }_{n}=\frac{R}{2L}$

A second-order highpass filter is implemented by taking the output of the filter to be the inductor voltage in the series RLC circuit. The resulting system function is

$\begin{array}{cc}\hfill {H}_{hp}\left(s\right)& =\frac{{V}_{hp}\left(s\right)}{{V}_{i}\left(s\right)}\hfill \\ & =\frac{{s}^{2}}{{s}^{2}+\frac{R}{L}s+\frac{1}{LC}}\hfill \end{array}$

So in terms of the undamped natural frequency and damping ratio, the second-order highpass filter is given by

${H}_{hp}\left(s\right)=\frac{{s}^{2}}{{s}^{2}+2\zeta {\Omega }_{n}s+{\Omega }_{n}^{2}}$

The graph of the frequency response magnitude of this filter is shown in [link] for several damping ratios and ${\Omega }_{n}=10$ . Frequency response of second-order highpass filter with Ω n = 10 and several values of ζ .

Not surprisingly, a second-order bandpass filter results by taking the output of the filter be the resistor voltage in the series RLC circuit. The system function is then

$\begin{array}{cc}\hfill {H}_{bp}\left(s\right)& =\frac{{V}_{bp}\left(s\right)}{{V}_{i}\left(s\right)}\hfill \\ & =\frac{\frac{R}{L}s}{{s}^{2}+\frac{R}{L}s+\frac{1}{LC}}\hfill \end{array}$

This can be expressed in terms of $\zeta$ and ${\Omega }_{n}$ as

${H}_{bp}\left(s\right)=\frac{2\zeta {\Omega }_{n}s}{{s}^{2}+2\zeta {\Omega }_{n}s+{\Omega }_{n}^{2}}$

## Frequency response of second-order bandpass filter

Setting $s=j\Omega$ in [link] gives the frequency response of the second-order bandpass filter

${H}_{bp}\left(j\Omega \right)=\frac{j2\zeta {\Omega }_{n}\Omega }{{\Omega }_{n}^{2}-{\Omega }^{2}+j2\zeta {\Omega }_{n}\Omega }$

The magnitude of this frequency response is shown in [link] where ${\Omega }_{n}=10$ . Frequency response of second-order bandpass filter with Ω n = 10 and several values of ζ .

Evidently, the frequency response peaks at $\Omega ={\Omega }_{n}$ . To prove this, we can divide the numerator and denominator of [link] by $j2\zeta {\Omega }_{n}\Omega$ , giving

$\begin{array}{cc}\hfill {H}_{bp}\left(j\Omega \right)& =\frac{1}{1-j\frac{{\Omega }_{n}^{2}-{\Omega }^{2}}{2\zeta {\Omega }_{n}\Omega }}\hfill \\ & =\frac{1}{1+j\frac{{\Omega }^{2}-{\Omega }_{n}^{2}}{2\zeta {\Omega }_{n}\Omega }}\hfill \\ \hfill & =\frac{1}{1+jA\left(\Omega \right)}\hfill \\ \hfill \end{array}$

where

$A\left(\Omega \right)=\frac{{\Omega }^{2}-{\Omega }_{n}^{2}}{2\zeta {\Omega }_{n}\Omega }$

It is easy to check that $\left|A,\left(,{\Omega }_{n},\right)\right|=0$ and $\left|A,\left(,\Omega ,\right)\right|>0,\Omega \ne {\Omega }_{n}$ . Therefore $\left|{H}_{bp},\left(j\Omega \right)\right|$ does in fact peak at $\Omega ={\Omega }_{n}$ .

The bandwidth of bandpass filters is typically measured as the difference between the two frequencies, ${\Omega }_{2}$ and ${\Omega }_{1}$ , at which the gain has dropped by 3 decibels from its peak of 0 dB. A diagram of this is shown in [link] . Definition of bandpass filter bandwidth, B W = Ω 2 - Ω 1 .

Formulas for ${\Omega }_{1}$ and ${\Omega }_{2}$ can be found by solving

$A\left(\Omega \right)=±1$

since if $A\left(\Omega \right)=±1$ , we get $\left|H,\left(,j,\Omega ,\right)\right|=\frac{1}{\sqrt{2}}$ . Solving the quadratic equation $A\left(\Omega \right)=1$ gives two solutions:

${\stackrel{˜}{\Omega }}_{1}=\zeta {\Omega }_{n}+{\Omega }_{n}\sqrt{1+{\zeta }^{2}}$
${\stackrel{˜}{\Omega }}_{2}=\zeta {\Omega }_{n}-{\Omega }_{n}\sqrt{1+{\zeta }^{2}}$

while solving $A\left(\Omega \right)=-1$ gives two additional solutions

${\stackrel{˜}{\Omega }}_{3}=-\zeta {\Omega }_{n}+{\Omega }_{n}\sqrt{1+{\zeta }^{2}}$
${\stackrel{˜}{\Omega }}_{4}=-\zeta {\Omega }_{n}-{\Omega }_{n}\sqrt{1+{\zeta }^{2}}$

Since ${\Omega }_{n}\sqrt{1+{\zeta }^{2}}>\zeta {\Omega }_{n}$ , we pick the two positive solutions as the “corner” frequencies:

${\Omega }_{1}=-\zeta {\Omega }_{n}+{\Omega }_{n}\sqrt{1+{\zeta }^{2}}$
${\Omega }_{2}=\zeta {\Omega }_{n}+{\Omega }_{n}\sqrt{1+{\zeta }^{2}}$

The two negative frequencies are just the negatives of the two positive frequencies (recall that $\left|H,\left(,j,\Omega ,\right)\right|$ has even symmetry). So the bandwidth is $BW={\Omega }_{2}-{\Omega }_{1}=2\zeta {\Omega }_{n}$ . It can also be readily verified that ${\Omega }_{n}=\sqrt{{\Omega }_{1}{\Omega }_{2}}$ , that is, the center frequency of the bandpass filter is the geometric mean of the two corner frequencies. The qualify factor , ${Q}_{o}$ is defined as

${Q}_{o}=\frac{{\Omega }_{n}}{BW}$

and is a measure of the narrowness of the filter bandwidth, with respect to its center frequency. It can be seen that ${Q}_{o}=\frac{1}{2\zeta }$ .

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