I attempted to design a low-pass 2nd Order Sallen key filter that exhibits a butterworth response. Unfortunately I'm getting confused in the math that describes the response. My design has a quality factor of approximately 0.707 and a cut-off frequency of 2kHz, which to my knowledge this is equivalent to a butterworth response. I looked up the tables for a 2nd order butterworth response and obtained the polynomial:


Does my pole frequency wn have to be equal to 1 for my design to exhibit a butterworth response? Thanks.


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  • 1
    \$\begingroup\$ No, that's the normalized response with wn = 1. They're designed that way using tables, then you just shift the frequency and can have all the poles. If your Q is 0.707 then it's a butter (for order 2). Check here en.wikipedia.org/wiki/… \$\endgroup\$ – Andrés Apr 10 '18 at 20:15
  • \$\begingroup\$ @Andrés Please do not answer questions in the comment section. It even says so when you start writing one. The user can't accept it if it's correct, so it breaks the whole idea with Stack Exchange. \$\endgroup\$ – pipe Apr 11 '18 at 6:16

Your sallen key filter has a gain of 1 hence it posseses this transfer function: -

\$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}\$

So, if \$\omega_n\$ (the natural resonant frequency) is normalized to 1 you get: -

\$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{s^2+2\zeta s+1}\$ where \$2\zeta = \dfrac{1}{Q}\$

If your Q = 0.707, the inverse is 1.414 (as seen in your polynomial).

It doesn't matter what value \$\omega_n\$ actually is; for a butterworth response Q = \$\dfrac{1}{\sqrt2}\$

Q is always the transfer function gain at \$\omega_n\$ and, for a butterworth response the gain at \$\omega_n\$ is always -3dB or 0.7071. This produces zero peaking in the pass-band i.e. is maximally flat in the pass-band: -

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Picture source

The natural resonant frequency (\$\omega_n\$) in your circuit is \$\dfrac{1}{\sqrt{R_1R_2C_1C_2}}\$.

I added this point just in case you were wondering about it.

The full TF of your circuit is: -

enter image description here

From here

  • \$\begingroup\$ I understand, its just normalized to 1. Thank you for your detailed answer. \$\endgroup\$ – Bonavia Apr 11 '18 at 18:15

No.. The various filter types are not determine by \$\omega\$ but rather by \$Q=\dfrac{1}{2\sigma}= 0.707= Butterworth\$

With higher order filters, if one allows more ripple in the passband from cascaded stagger peaks, you can get steeper skirts before -6dB/octave per order.

Filter type        3rd Max Q   2nd Max Q  1st Order
-----------        --------    ---------  --------
Butterworth        Q=1.0        Q=0.71     Q=0.5
Bessel             Q=0.69       Q=0.58   
Linear Phase 0.05° Q=0.80       Q=0.60  
Linear Phase 0.5°  Q=0.95       Q=0.64 
Chebychev 0.5dB    Q=1.706      Q=0.864  
Chebychev 0.957dB  Q=2.018      Q=0.957
Gaussian to 6dB    Q=0.81       na
Gaussian to 12dB   Q=0.82       na

As the order increases in Buttworth the group delay at the breakpoint gets really huge , where as Bessel is maximally flat group delay so it has lower Q.

Chebychev Filters on the otherhand, have much high Q's with steeper attenuation on the "skirts" but higher Q demands tighter tolerances.

e.g. here using 0.5% tolerances in 10th order that could be RRIO Quad Op Amps

enter image description here

Can you see where the Q=0.707 comes from, with the 1.414 "s" term?


For a 2nd order filter, it would be nice to allow the gain, frequency, and damping to be independently settable. This puts restrictions on the gain and the ratios of the resistors and capacitors you use.

For some frequency, \$\omega\$, such a low pass filter equation looks like:

$$\begin{align*}\frac{e_\text{OUT}}{e_\text{IN}} &=\frac{K\:\omega^2}{s^2+d\:\omega\:s +\omega^2}\label{eq1}\tag{Two Pole Low Pass}\end{align*}$$

For analysis, \$\omega=1\$ and gain \$K=1\$, so:

$$\begin{align*}\frac{e_\text{OUT}}{e_\text{IN}} &=\frac{1}{1 +d\:s +s^2}\label{2PLP}\tag{Two Pole Analysis}\end{align*}$$

The above \$\ref{2PLP}\$ equation is the single equation representing every single useful two-pole low-pass filter. (Note that \$d\$ is the damping value and not the damping ratio. Be careful here, because most modern teaching is done using the damping ratio.)

A Butterworth has \$d=\sqrt{2}\$. So the above \$\ref{2PLP}\$ equation set for a Butterworth filter, with \$\omega=1\$, is:

$$\begin{align*}\frac{e_\text{OUT}}{e_\text{IN}} &=\frac{1}{1 +\sqrt{2}\:s +s^2}\label{2PLPA}\end{align*}$$

But that's for \$\omega=1\$.

For the Sallen-Key arrangement, \$\omega^2=\frac{1}{R_1\:C_1\:R_2\:C_2}\$, and the resulting equation is:

$$\begin{align*}\frac{e_\text{OUT}}{e_\text{IN}} &=\frac{\frac{K}{R_1\:C_1\:R_2\:C_2}}{s^2+\left(\frac{1}{R_1\:C_1}+\frac{1}{R_2\:C_1}+\frac{1-K}{R_2\:C_2}\right)\:s +\frac{1}{R_1\:C_1\:R_2\:C_2}}\end{align*}$$

So, given your circuit with \$K=1\$, I get about \$f=1927\:\text{Hz}\$ and \$d=1.4071\$. That's pretty close to a Butterworth filter. (Of course, parts have accuracy bars to them; not accounted for here.)

\$\omega=1\$ for analysis purposes. In that case, the transfer function is the above \$\ref{2PLP}\$ equation. However, you then translate that into the transfer function at some \$\omega\$ as shown in \$\ref{eq1}\$ equation. (Set \$K=1\$ for your topology.)

The analysis equation is designed to expose the damping value as a new addition to the one-pole filter (which lacks it.) All 2-pole low pass filters can be analyzed using it, keeping in mind that \$\omega=1\$ at the time. So that's why you see the value \$d=1.414\$ for a Butterworth. They are presuming that you understand that this is in the context of \$\omega=1\$ and that you know how to shift the values to get the cutoff frequency you want while keeping the damping factor the same. You can set these details independently. Which is an important facet of this process.


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