I'm trying to learn more about high order filter design, with my goal being to design an 8th order Butterworth lowpass filter in spice, then in the real world. I've been reading a filter book I have and I was trying to follow this tutorial here.

I was doing ok up until this point:

From the normalised low pass Butterworth Polynomials table above, the coefficient for a third-order filter is given as \$(1+s)(1+s+s^2)\$ and this gives us a gain of \$3-A = 1\$, or \$A = 2\$. As \$A = 1 + (R_f/R_1)\$, choosing a value for both the feedback resistor \$R_f\$ and resistor \$R1\$ gives us values of \$1\$kΩ and \$1\$kΩ respectively, ( \$1\$kΩ\$/1\$kΩ\$ + 1 = 2\$ ).

I don't understand how they went from \$(1+s)(1+s+s^2)\$, and all of a sudden they know the gain from that which you use to calculate your resistors. There's no more explanation, this basically reads to me as if you start with this polynomial table, then a wizard appears, and now you know your gain.

So how did they go from the polynomial to the gain? The 8th order equation is even larger:


Does that imply that I will have different gain settings for each stage?

  • 1
    \$\begingroup\$ I THINK it implies an assumption that the gain will be 1 (or 2 in the case of a non inverting stage). I'm definitely not an expert in the math and will often resort to "cookbook" equations myself, but from what I understand adding gain beyond these baseline "norms" adds complications to the equations which are normally ignored, because they factor out when there is no added gain. (not to mention adding often undesired peaks in the response curves). \$\endgroup\$
    – Randy
    Nov 21, 2014 at 17:23

2 Answers 2


@Confused: Sorry - but I have to start with some general comments:

In principle, for an 8th order filter you have two basic alternatives:

(a) Direct realization (active topology derived from a passive and tabulated reference structure, and (b) Cascade realization as a series connection of 4 second-order stages.

I suppose, you are following the latter approach - and here you have again several alternatives (how the various 2nd order stages are realized). It seems that you have decided to use Sallen-Key realizations because you have mentioned finite gain values.

But also in this case, you again have alternatives: Unity gain approach, gain-of-two approach or equal-component approach (with gain values lower than "3"). Independent on these 3 alternatives, you must know that all 4 stages look different: They are individually designed for equal pole frequencies (applies only for Butterworth response) but for different pole-Q values to be found in filter tables. Hence, you will NOT have 4 identical 2nd-order stages but each of the 4 must bedesigned separately.

I am not sure if this answers all of your questions - perhaps it helps if you could give us some more detail of your envisaged design.

EDIT 1: The following link leads you to a document (from TI) which gives you the Q values for your 8th-order filter on page 8 (correction: page 9)


EDIT 2: For your convenience, here are the formulas for designing the 4 stages (equal pole frequencies wp, different Qp values) - to be applied for the gain-of-two version:

C1=C2=C and wp=1/[C*Sqrt(R1R3)] and Qp=Sqrt(R1/R3) with R1: Most left resistor(connected to input signal). For a gain of "2" you can use any two equal resistors in the negative feedback path.

  • \$\begingroup\$ Thank you that clears up somethings for me, I have read although not fully understood that TI note. The design idea I had was just 4 cascaded stages for an 8th order Butterworth, ideally with no gain. I'm having trouble understanding how to design each stage, and how to go from these polynomial equations to values for R and C. \$\endgroup\$
    – confused
    Nov 21, 2014 at 18:53
  • \$\begingroup\$ I gave you some hints in the other thread about Butterworth response. Corresponding formulas which relate component values to pole data can be found in several publications. If you have problems, I can help you. However, it is necessary to decide which Sallen-Key structure you are goig to realize. \$\endgroup\$
    – LvW
    Nov 21, 2014 at 21:11
  • \$\begingroup\$ I like to add that I do NOT recommend to use the unity-gain version of the Sallen-Key topology for realizing an 8th order filter. Calculation/selection of parts is more involved if compared with the gain-of-two version (which is less-known). In this version you may use two identical capacitors. \$\endgroup\$
    – LvW
    Nov 22, 2014 at 8:22
  • \$\begingroup\$ Further details can be found in my above answer (EDIT 2). \$\endgroup\$
    – LvW
    Nov 22, 2014 at 8:58
  • \$\begingroup\$ So I went back to basics, read up on Laplace transforms, poles and zeros, then Saleen key structure, and then back to Butterworth. I looked again at the TI application note, and I think I understand the method they are using in section 10. So I spice their filter in this example and it does come out to 1K at -3dB. I do the math along with the problem and it adds up. Then I try to design a filter for 5MHz. I keep the same m and n values they used, and using the formulas I get R1=2.29K, R2=10K, C1=3.6pF, and C2=11.88pF. \$\endgroup\$
    – confused
    Nov 22, 2014 at 23:04

So how did they go from the polynomial to the gain?

They didn't go from the polynomial directly to the gain. Your reference link actually omit something, then come from other place.

The equation 3 - A = 1 is related to the stability of the filter. It actually applies to second-order Equal-Component KRC filter (An implementation of Sallen-Key filter, with equal resistors and equal capacitors). For this type of filter the

$$ Q = \frac{1}{3-K} \quad\quad\quad\quad\quad\quad\quad(1) $$

\$K\$ is the DC gain of the filter, you see if K is equal to or greater than 3, the Q will be infinite, event be negative! This will make the filter unstable. So, K must be less than 3. That is, the system should have DC gain less then 3.

So the author intends to use this type of filter structure to make a Butterworth filter, and forgets to mention this detail.

In a second-order system, the damping factor has a relation with Q

$$ Q = \frac{1}{2\zeta}\quad\quad\quad\quad\quad\quad\quad\quad(2) $$

So, combine (1) and (2)

$$ K=3-2\zeta $$

As we known, \$\zeta\$ should be between 0 ~ 1. The author choose 3 - A = 1, he actually set the \$\zeta\$ to 0.5, and then the gain should equal to 2. Because he choose the equal-component KRC implementation, so \$R_{f} = R_{1}\$, and the gain = 1+Rf/R1, just equal to 2. Then all are perfect.

  • \$\begingroup\$ ....perfect? I think, for ζ=0.5 we have Qp=1 which does NOT give a Butterworth response. Instead, we need Qp=0.7071. \$\endgroup\$
    – LvW
    Nov 22, 2014 at 13:48
  • \$\begingroup\$ Um, maybe something is wrong... \$\endgroup\$
    – diverger
    Nov 22, 2014 at 14:13
  • \$\begingroup\$ @LvW: I think this is just what the author want to do. Perfcet means we got the author's idea. \$\endgroup\$
    – diverger
    Nov 22, 2014 at 14:28
  • \$\begingroup\$ @LvW: I think the author made a mistake. \$\endgroup\$
    – diverger
    Nov 22, 2014 at 14:36
  • \$\begingroup\$ No, the didn`t a mistake. For a third-order Butterworth filter the active stage of 2nd order must have - indeed - a pole-Q of "1". However - in our case, the questioner needs n=8, and this requires other pole-Q values for each stage (to be found in the referenced TI document). \$\endgroup\$
    – LvW
    Nov 22, 2014 at 16:48

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