# What is the $Q$ of successive Second-Order Sections of an $N$th-order Butterworth low-pass filter?

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?

• I plan to answer this myself unless someone else beats me to it. I'll let it hang for a day. Commented Jul 25, 2021 at 3:16
• I would draw the Butterworth circle, take the stable poles, divide those into pairs, and come up with a formula, what do you think? Commented Jul 25, 2021 at 7:46
• 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). Commented Jul 25, 2021 at 11:57
• 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 $N$. Commented Jul 25, 2021 at 20:37
• You still didnt say what is the use of all of this (: Commented Jul 27, 2021 at 6:41

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.

• I upvoted Robert's answer, too, but yours is just what I would have also said. Commented Jul 28, 2021 at 14:25
• 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. Commented Jul 28, 2021 at 15:40
• Why are our results different, Matt? Commented Jul 28, 2021 at 17:51
• @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? Commented Jul 28, 2021 at 17:58
• @robertbristow-johnson: That's ok with me, even though I think that readers won't get too distracted by the slight differences in notation. Commented Jul 28, 2021 at 20:59

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

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