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What if you had an $N$th-order Butterworth lowpass filter filter with -3 dB frequency of $\Omega_0$?

$$ \Big| H(j\Omega) \Big|^2 = \frac{1}{1 + \left(\frac{\Omega}{\Omega_0}\right)^{2N}} $$

The number of second-order sections (SOS) or biquads in series is $\left\lfloor \frac{N}{2} \right\rfloor$. We know that the resonant frequency for each SOS is $\Omega_0$.

If the order $N$ is even:

$$ H(s) = \prod\limits_{n=1}^{\frac{N}{2}} \frac{1}{1 + \frac{1}{Q_n}\frac{s}{\Omega_0} + \left(\frac{s}{\Omega_0}\right)^2} $$

If the order $N$ is odd:

$$ H(s) = \frac{1}{1 + \frac{s}{\Omega_0}} \ \prod\limits_{n=1}^{\frac{N-1}{2}} \frac{1}{1 + \frac{1}{Q_n}\frac{s}{\Omega_0} + \left(\frac{s}{\Omega_0}\right)^2} $$

What is the $Q_n$ for each second-order section?

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  • $\begingroup$ I plan to answer this myself unless someone else beats me to it. I'll let it hang for a day. $\endgroup$ Commented Jul 25, 2021 at 3:16
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    $\begingroup$ I would draw the Butterworth circle, take the stable poles, divide those into pairs, and come up with a formula, what do you think? $\endgroup$
    – Ran Greidi
    Commented Jul 25, 2021 at 7:46
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    $\begingroup$ The individual $Q$'s are determined by the real parts of the poles, which lie on a circle. So I guess it's a pretty straightforward result, something like $$Q_n=-\frac{1}{2\cos(\pi (2n+N-1)/2N)}$$ (didn't take time to check the details, so don't hold me to it, it's just a comment after all). $\endgroup$
    – Matt L.
    Commented Jul 25, 2021 at 11:57
  • $\begingroup$ both @RanGreidi and Matt L got it, i think. i thought i was gonna have two slightly different formulae for even $N$ vs. odd $N$ but maybe if you start out on the "Butterworth circle" next to the $j\Omega$ axis and proceed toward the negative real axis, you get a consistent formula that is the same for even or odd $N$. $\endgroup$ Commented Jul 25, 2021 at 20:37
  • $\begingroup$ You still didnt say what is the use of all of this (: $\endgroup$
    – Ran Greidi
    Commented Jul 27, 2021 at 6:41

2 Answers 2

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It is well-known that the poles of a normalized continuous-time $N^{th}$-order Butterworth lowpass filter lie on a semi-circle with radius $1$, centered at $s=0$:

$$p_n= e^{j\pi(2n-1+N)/2N},\qquad n=1,\ldots,N\tag{1}$$

Note that for odd order $N$, there is a single pole at $s=-1$.

Combining the complex conjugate pole pairs, we can construct a polynomial

$$D(s)=\prod_{n=1}^{\left\lfloor\frac{N}{2}\right\rfloor}(s-p_n)(s-p_n^*)=\prod_{n=1}^{\left\lfloor\frac{N}{2}\right\rfloor}\big(s^2-2\textrm{Re}\{p_n\}s+1\big)\tag{2}$$

Using $(1)$, the polynomial $D(s)$ can be written as

$$D(s)=\prod_{n=1}^{\left\lfloor\frac{N}{2}\right\rfloor}\left[s^2-2\cos\left(\frac{\pi(2n-1+N)}{2N}\right)s+1\right]\tag{3}$$

For even $N$, the denominator of the filter's transfer function $H(s)$ equals $D(s)$, and for odd $N$ we have another factor due to the single pole at $s=-1$:

$$H(s)=\begin{cases}\displaystyle\frac{1}{D(s)},&N\textrm{ even}\\\displaystyle\frac{1}{(1+s)D(s)},&N\textrm{ odd}\end{cases}\tag{4}$$

Consequently, for even as well as for odd orders, the $Q$ factors of the individual second-order sections are given by

$$Q_n=-\frac{1}{2\cos\left(\frac{\pi(2n-1+N)}{2N}\right)},\qquad n=1,\ldots,\left\lfloor\frac{N}{2}\right\rfloor\tag{5}$$

Or, with $\cos(x+\pi/2)=-\sin(x)$, Eq. $(5)$ can also be written as

$$Q_n=\frac{1}{2\sin\left(\frac{\pi}{N}\left(n-\frac12\right)\right)},\qquad n=1,\ldots,\left\lfloor\frac{N}{2}\right\rfloor\tag{6}$$

which agrees with the result in Robert's answer.

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    $\begingroup$ I upvoted Robert's answer, too, but yours is just what I would have also said. $\endgroup$ Commented Jul 28, 2021 at 14:25
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    $\begingroup$ yeah i was gonna show the placements of the poles and derive it. but finishing these answers rigorously takes more time than i expect. The purpose of the question was to establish a closed-form expression that people can couple to the Cookbook and whip out a quick design. $\endgroup$ Commented Jul 28, 2021 at 15:40
  • $\begingroup$ Why are our results different, Matt? $\endgroup$ Commented Jul 28, 2021 at 17:51
  • $\begingroup$ @robertbristow-johnson: They're the same, I guess. I just didn't remove the $\pi/2$ from the argument of the cosine. My result is the same as your last line before you change from cosine to sine, ain't it? $\endgroup$
    – Matt L.
    Commented Jul 28, 2021 at 17:58
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    $\begingroup$ @robertbristow-johnson: That's ok with me, even though I think that readers won't get too distracted by the slight differences in notation. $\endgroup$
    – Matt L.
    Commented Jul 28, 2021 at 20:59
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Okay, since

$$ H(j \Omega) = H(s) \Big|_{s=j\Omega} $$

then

$$\begin{align} \Big| H(j\Omega) \Big|^2 &= \frac{1}{1 + \left(\frac{\Omega}{\Omega_0}\right)^{2N}} \\ \\ \Big| H(s) \Big|^2 &= \frac{1}{1 + \left(\frac{s}{j\Omega_0}\right)^{2N}} \\ \\ \end{align} $$

Poles, $p_n$, occur at values of $s$ where the denominator goes to zero.

$$\begin{align} 1 + \left(\tfrac{s}{j\Omega_0}\right)^{2N} \Bigg|_{s = p_n} &= 0 \\ \\ 1 + \left(\tfrac{p_n}{j\Omega_0}\right)^{2N} &= 0 \\ \\ \left(\tfrac{p_n}{j\Omega_0}\right)^{2N} &= -1 \\ \\ \left(\tfrac{p_n}{j\Omega_0}\right)^{2N} &= \underbrace{e^{-j\pi}}_{-1} \ \underbrace{e^{j 2\pi n}}_{1} \qquad n \in \mathbb{Z} \\ \\ \left(\tfrac{p_n}{j\Omega_0}\right)^{2N} &= e^{j\pi(-1 + 2n)} \\ \\ \frac{p_n}{j\Omega_0} &= e^{j\pi(2n-1)/(2N)} \\ \\ p_n &= j\Omega_0 e^{j\pi(2n-1)/(2N)} \\ \\ &= e^{j \frac{\pi}{2}} \Omega_0 e^{j\pi(2n-1)/(2N)} \\ \\ &= \Omega_0 \ e^{j \frac{\pi}{2}} \ e^{j\pi(\frac{n}{N}-\frac{1}{2N})} \\ \\ \end{align} $$

These are all on a circle in the $s$-plane of radius $\Omega_0$

The $N$ in the left-half $s$-plane are the ones we use and they correspond to $1 \le n \le N$.

When $N$ is even, all poles are complex conjugate pairs.

When $N$ is odd, the pole corresponding to $n = \frac{N+1}{2}$ is a single real pole located at $p_n = -\Omega_0$. All other poles are complex conjugate pairs.

$$ p_{N+1-n} = \big( p_n \big)^* $$

So, each second-order section (SOS) in the product is

$$\begin{align} \frac{1}{1 + \frac{1}{Q_n}\frac{s}{\Omega_0} + \left(\frac{s}{\Omega_0}\right)^2} &= \frac{\Omega_0^2}{s^2 + \frac{\Omega_0}{Q_n}s + \Omega_0^2} \\ &= \frac{\Omega_0^2}{(s-p_n)(s-p_n^*)} \qquad \qquad \text{for } 1 \le n \le \left\lfloor \tfrac{N}{2} \right\rfloor \\ &= \frac{\Omega_0^2}{s^2 - (p_n+p_n^*)s + p_n p_n^*} \\ \\ &= \frac{\Omega_0^2}{s^2 - 2 \Re e \{p_n\}s + |p_n|^2} \\ \end{align}$$

This results in:

$$ |p_n| = \Omega_0 $$

$$\begin{align} \frac{\Omega_0}{Q_n} &= -2 \Re e \big\{ p_n \big\} \\ \\ &= -2 \Re e \Big\{\Omega_0 \ e^{j \frac{\pi}{2}} \ e^{j\pi(\frac{n}{N}-\frac{1}{2N})} \Big\} \\ \\ &= -2\Omega_0 \ \Re e \Big\{ e^{j\pi(\frac{1}{2} + \frac{n}{N} -\frac{1}{2N})} \Big\} \\ \\ &= -2\Omega_0 \ \cos\big( \pi(\tfrac{1}{2} + \tfrac{n}{N} -\tfrac{1}{2N}) \big) \\ \\ &= -2\Omega_0 \ \cos\big( \tfrac{\pi}{2} + \pi\tfrac{n}{N} -\tfrac{\pi}{2N} \big) \\ \\ &= 2\Omega_0 \ \sin\big( \tfrac{\pi}{N}(n -\tfrac{1}{2}) \big) \\ \end{align}$$

So, it appears to me that

$$ Q_n = \frac{1}{2 \sin\big( \tfrac{\pi}{N}(n -\tfrac{1}{2}) \big)} \qquad \qquad \text{for } 1 \le n \le \left\lfloor \tfrac{N}{2} \right\rfloor $$

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