I need to design a Butterworth band-pass filter with N = 2 with central frequency of \$f_0\$, bandwith of \$\Delta\$ and gain \$G\$.

I tried to make sense of it mathematically. So I first derived the equation for a band-pass filter

$$T(jw) = K\frac{\frac{1}{Q}\omega_0s}{s^2+\frac{1}{Q}\omega_0s+\omega_0^2} = K\frac{Bs}{s^2+Bs+\omega_0^2}$$

In the last passage I substituted \$B = \frac{\omega_0}{Q}\$.

With some manipulation (and considering \$Q^{-2} - 2 = 0\$) I got

$$ |T_n(jw)| = \frac{\sqrt{K^2\frac{1}{Q^2}\frac{\omega^2}{\omega_0^2}}}{\sqrt{\left(1-\frac{\omega^2}{\omega_o^2}\right)^2 + \frac{1}{Q^2}\frac{\omega^2}{\omega_0^2}}} =\frac{K\frac{1}{Q}(\frac{\omega}{\omega_0})}{\sqrt{1 + \left(\frac{\omega}{\omega_0}\right)^{4}}} $$

My first question arises here. Shouldn't I get a Butterworth filter of \$N = 4\$ and not \$N=2\$?

I saw the solution to the problem and what the textbook did was, because it is a passband, to consider \$N=1\$. But there's no explanation whatsoever. So this is my second question: why can we consider \$N = 1\$ when asked for a \$N = 2\$ filter?

It then followed a Butterworth table for which at \$N = 1\$ the transfer function would have the form \$T(S) = \frac{1}{S + 1}\$. And then defines \$S = \epsilon^{1/N}(\frac{s^2+\omega_0^2}{Bs})\$ for which it assumes \$epsilon = 1\$. What is this \$epsilon\$ and why should we consider it 1 in this case? It also references that the gain at natural frequency is \$3 dB\$.

  • 2
    \$\begingroup\$ I know, there is a slight confusion regarding the filter order N for bandpass circuits. As far as I know, most books and other knowledge sources consider N=2 as the smallest order for bandpass functions (because - according to the lowpaas-bandpass transformation - each 1st order lowpass function always is transferred to a 2nd-order bandpass function, which therefore is the smallest order. This is in full accordance with the denominator which also is ALWAYS at least of the order N=2 for a bandpass. \$\endgroup\$
    – LvW
    Commented Oct 14, 2022 at 17:10
  • 2
    \$\begingroup\$ You don't get to specify the bandwidth for a 2nd order Butterworth. The shape of the bandwidth response is already determined by the fact that a 2nd order Butterworth has \$Q=\frac{\sqrt{2}}{2}\$. Butterworth hard-codes that shape for you. (And in any case the low and high frequencies are both imaginary values in this underdamped case so the meaning of bandwidth is a little shaky -- or imaginary.) You do get to specify \$\omega_{_0}=2\pi\, f_{_0}\$, though. \$\endgroup\$
    – jonk
    Commented Oct 14, 2022 at 18:40
  • 1
    \$\begingroup\$ In answer to your first question, your result where the 4th power appears in your RHS doesn't mean \$N=4\$ The magnitude multiplies two 2nd order factors and it gets, of course, a 4th order result. But you also take the square-root of it. So that brings it back to 2nd order, if you want think about it that way. \$\endgroup\$
    – jonk
    Commented Oct 14, 2022 at 18:41
  • \$\begingroup\$ @jonk this may come as a silly question, but I thought they were the same (just different notation). What is the difference between \$f_x\$ and \$\omega_x\$? Why can I change one but not the other for a Butterworth filter? \$\endgroup\$
    – ludicrous
    Commented Oct 15, 2022 at 10:11
  • \$\begingroup\$ @ludicrous I only wrote \$\omega_{_0}=2\pi\,f_{_0}\$. That's a relation or, I suppose, a change of variable. But I never suggested (or, at least, wanted to suggest) that they are independent of each other. People use one or the other and it's a good idea to keep in mind their relationship when interpreting communications. Did I write poorly, above? \$\endgroup\$
    – jonk
    Commented Oct 15, 2022 at 21:47

1 Answer 1


First, both @LvW and @jonk are right in the comments: the order for bandpass/bandstop filters are a bit of a mess, and for a 2nd order transfer function that is a bandpass (your first formula), the \$Q\$ dictates the bandwidth so, if you set it to be \$\frac{1}{\sqrt2}\$ then the -3 dB edges will be at \$\frac{1}{\sqrt2}\$ Hz and \$\sqrt2\$ Hz (relative to 1 Hz center frequency).

However, further on you say two things:

  1. you expected to have a resulting 4th order transfer function
  2. you're wondering what the \$\epsilon\$ (\epsilon for MathJax) is

For 1. it's simple: start from the lowpass prototype and apply the frequency transformation:

$$H_{\text{LP}}(s)\rightarrow H_{\text{BP}}\left(\dfrac{s^2+\omega^2}{BW s}\right) \tag{1}$$

Funny enough, today I wrote this answer where I show the meaning of that transformation:

$$\dfrac{s^2+\omega^2}{BW s}=\dfrac{s}{BW}+\dfrac{1}{s\frac{BW}{\omega^2}} \tag{2}$$

So now apply (1) and you'll get the transfer function of the bandpass to be made of a lowpass and a highpass:

$$\begin{align} H(s)&=\dfrac{\omega_0^2}{s^2+\dfrac{\omega_0}{Q}s+\omega_0^2} \\ H\left(\dfrac{s^2+\omega^2}{BW s}\right)&=\dfrac{BW^2\omega_0^2s^2}{s^4+\dfrac{\omega_o}{Q}BWs^3+(2\omega_c^2+BW^2\omega_0^2)s^2+\dfrac{\omega_0}{Q}BW\omega_c^2s+\omega_c^4} \tag{3}\\ &=\dfrac{\omega_1^2}{s^2+\dfrac{\omega_1}{Q_1}s+\omega_1^2}\cdot\dfrac{s^2}{s^2+\dfrac{\omega_2}{Q_2}s+\omega_2^2} \tag{4} \end{align}$$

Or, if you started from the poles of the lowpass prototype, convert them directly to bandpass with the following (which can be obtaned by solving for the roots of the denominator of (3)):

$$\begin{align} s_{_\text{LP}}&=\sigma\pm j\omega \\ M&=|s_{_\text{LP}}| \\ \Phi&=\angle{s_{_\text{LP}}} \\ &\begin{cases} s{1,2}=\dfrac12\left\{-BW\sigma+\sqrt{M}\cos\left(\dfrac{\Phi}{2}\right)\pm j\left[BW\omega-\sqrt{M}\sin\left(\dfrac{\Phi}{2}\right)\right]\right\} \\ s{3,4}=\dfrac12\left\{-BW\sigma-\sqrt{M}\cos\left(\dfrac{\Phi}{2}\right)\pm j\left[BW\omega+\sqrt{M}\sin\left(\dfrac{\Phi}{2}\right)\right]\right\} \end{cases} \tag{5} \end{align}$$

Either way, you'll end up with (4), which is a 4th order transfer function which is a bandpass, or a 2nd order bandpass (implying a multiple a 2nd orders). Yes, it's confusing but, the most certain way to be sure about it is to specify it as a 4th order transfer function. This way, you are certain that the denominator has an \$s^4\$, no more, no less.

And about 2., this is related to the generic way in which the Butterworth mathematical transfer function is written:

$$\begin{align} H(s)&=\dfrac{1}{\sqrt{1+\epsilon^2\left(\dfrac{\omega}{\omega_0}\right)^{2N}}} \tag{6} \\ \epsilon&=\sqrt{10^{\frac{A_{\text{dB}}}{10}}-1} \end{align}$$

That \$\epsilon\$ dictates the attenuation at \$\omega_0\$. If the attenuation is to be the "classical" -3(.0103...) dB then:

$$\begin{align} A_{\text{dB}}&=-20\log_{10}\left(\dfrac{1}{\sqrt2}\right)\approx -3\;\text{dB} \\ \epsilon&=\sqrt{10^{\frac{3}{20}}-1}=1 \tag{7} \end{align}$$

And, since in the majority of the cases \$\epsilon\$ doesn't need tinkering, it's considered 1. But nobody says that the -3 dB value is set in stone -- that is up to the requirements for the filter. For example, if the attenuation at \$\omega_0\$ needs to be 0.9 (0.915 dB) then \$\epsilon=0.4843...\$.

It puzzles me that it appears with a \$\frac1N\$ as power, in the OP. it shouldn't since (6) shows it to be squared, not a power of N. But they probably use a different notation an use it to extract the custom attenuation, which is:

$$\omega_x=\dfrac{\omega_0}{\left(\sqrt{10^{\frac{A_{\text{dB}}}{10}}-1}\right)^\frac1N}\quad\left[=\dfrac{\omega_0}{\epsilon^\frac1N}\right] \tag{8}$$

At any rate, the -3 dB convention is (mathematically) only valid for 1st orders and Butterworth filters, since that is the point that coincides with the phase being half its final (asymptotic) value. But, ultimately, the design dictates what frequency scaling needs to be applied and where (in most cases that's not needed).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.