# Multiple feedback design methodology

I am having trouble finding a step by step approach to the MFB filter. I came across a tutorial from Analog Devices but I am struggling to find what is the reasoning behind their equations, and can't find any references about it. Their design goes as this:

Now I understand there are two conditions that need to be imposed and then the other three components are determined (knowing the coefficients of the transfer function). I can't figure out what their reasoning is for deriving each equation and been around it for hours. I did determine the equations of the parameters:

$$H=\frac{R_4}{R_1}$$

$$\omega_0 = \displaystyle \frac{1}{\displaystyle \sqrt{R_3R_4C_2C_5}}$$

$$Q = \displaystyle \frac{\displaystyle \frac{1}{\displaystyle \sqrt{R_3R_4}}}{\displaystyle \frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}} \times \sqrt{\frac{C_2}{C_5}}$$

## To begin

Two equivalent forms of the low-pass standard form term (three terms are possible for the full 2nd order TF):

\begin{align*} \frac{V_{_\text{OUT}}}{V_{_\text{IN}}}&= A_0\cdot\left\{\begin{array}{l} \frac1{ \left(\frac{s}{\omega_{_0}}\right)^2+\frac1{Q} \left(\frac{s}{\omega_{_0}}\right)+1} \\\\ \frac{\omega_{_0}^2}{s^2+2\zeta\: \omega_{_0}s+\omega_{_0}^2} \end{array}\right. \end{align*}

(I use $$\A_0\$$ for the gain and not $$\H\$$, with the 0 implying the low-pass gain. A full 2nd order transfer function may also have $$\A_2\$$ and $$\A_1\$$ for the high-pass and band-pass terms.)

## This configuration

The multifeedback solution for your schematic is:

\begin{align*}A_0&=-\frac{R_4}{R_1}\\\\\omega_{_0}&=\frac1{\sqrt{R_3\,R_4\,C_2\,C_5}}\\\\\zeta&=\frac12\cdot\sqrt{\frac{C_5}{C_2}}\cdot\frac{\sqrt{R_3\,R_4}}{R_1\,\mid\mid\, R_3\,\mid \mid\, R_4}, \text{ or:}&Q&=\sqrt{\frac{C_2}{C_5}}\cdot\frac{R_1\,\mid\mid\, R_3\,\mid \mid\, R_4}{\sqrt{R_3\,R_4}}\end{align*}

## Sanity check

The form they used is closer to the $$\\zeta\$$-form, so I'd start with $$\2\zeta\,\omega_{_0}=\frac1{C_2}\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\$$. Since $$\2\pi \,f_{_0}=\omega_{_0}\$$ then their $$\k=\omega_{_0}\cdot C_5\$$. Also, their $$\\alpha=2\zeta\$$.

So let's start at the beginning and just plug things in for a sanity test:

\begin{align*} 2\zeta\,\omega_{_0}&=\frac1{C_2}\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right) \\\\&= \frac1{\frac4{\alpha^2}\left(1+H\right)C_5}\left(\frac1{\frac{\alpha}{2 \,H\, k}}+\frac1{\frac{\alpha}{2\,\left(1+H\right)\,k}}+\frac1{\frac{\alpha}{2\,k}}\right) \\\\&= \frac{\alpha^2}{4\left(1+H\right)C_5}\left(\vphantom{\frac1{\frac{\alpha}{2 \,H\, k}}}2 \,H+2\,\left(1+H\right)+2\right)\frac{k}{\alpha} \\\\&= \frac{\alpha^2}{4\left(1+H\right)C_5}\left(\vphantom{\frac1{\frac{\alpha}{2 \,H\, k}}}4 \,H+4\right)\frac{k}{\alpha} \\\\&= \frac{\alpha\,k}{C_5} \\\\&= \frac{\left(2\zeta\right)\,\left(\omega_{_0}\,C_5\right)}{C_5} \\\\&= 2\zeta\,\omega_{_0}\end{align*}

So their algorithm checks out. Someone sane was involved. Good.

## Understanding their algorithm

Now let's see how it might have been developed in the first place.

If we are given $$\C_5\$$, then it follows that $$\\frac1{C_2}=R_3\,R_4\,C_5\,\omega_{_0}^2\$$. So, using the definition for $$\\omega_{_0}\$$ we can find that:

\begin{align*} 2\zeta\,\omega_{_0}&= \frac1{C_2}\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\\\\ 2\zeta\,\omega_{_0}&=R_3\,R_4\,C_5\,\omega_{_0}^2\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\\\\ 2\zeta&=R_3\,R_4\,C_5\,\omega_{_0}\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\\\\ \frac{2\zeta}{\omega_{_0}\,C_5}&=R_3\,R_4\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\\\\ \frac{\zeta}{\omega_{_0}\,C_5}&=\frac12 R_3\,R_4\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right) \\\\ \text{Clearly, }\frac{\zeta}{\omega_{_0}\,C_5}&\text{ is a resistance -- substitute in the gain, now}\\\\&=\frac12 R_3\left(\frac{R_4}{R_1}+\frac{R_4}{R_3}+1\right)\\\\ &=\frac12 R_3\left(H+1+\frac{R_4}{R_3}\right)\\\\\text{To resolve }&\frac{R_4}{R_3},\text{ assign }R_4=\frac{\zeta}{\omega_{_0}\,C_5}\text{ and place it on the left side}\\\\ R_4&=\frac12 R_3\left(H+1+\frac{R_4}{R_3}\right)\\\\ \frac{R_4}{R_3}&=\frac12 \left(H+1+\frac{R_4}{R_3}\right)\\\\ \text{Then assign }&u\text{ as the ratio }\frac{R_4}{R_3}\text{ and solve}\\\\ u &= \frac12 \left(H+1+u\right)\\\\ 2 u &= H+1+u\\\\ u &= H+1 \end{align*}

Given the definition $$\R_4=\frac{\zeta}{\omega_{_0}\,C_5}=\frac1{2 Q\,\omega_{_0}\,C_5}\$$ and from the gain $$\\mid A_{_0}\!\mid=H=\frac{R_4}{R_1}\$$ requiring $$\R_1=\frac1{H}R_4\$$, it follows that remaining conditions then require $$\R_3=\frac1{H+1}R_4\$$.

So the resistors are done. (All this started out given $$\Q\$$, $$\\omega_{_0}\$$, and $$\C_5\$$. Nice.)

Now for $$\C_2\$$.

(I'm taking short-cuts, now, instead of writing out lists of tiny incremental steps I'm sure you can handle from here out.)

Return to $$\2\zeta\,\omega_{_0}=\frac1{C_2}\left(\frac1{R_1}+\frac1{R_3}+\frac1{R_4}\right)\$$. With the new resistor values, this becomes $$\2\zeta\,\omega_{_0}=\frac1{C_2}\cdot 2\left(H+1\right)\cdot\frac{\omega_{_0}\,C_5}{\zeta}\$$.

Solving, and canceling stuff out, find that $$\C_2=\frac{H+1}{\zeta^2}C_5\$$.

Substituting in $$\\zeta=\frac1{2\,Q}\$$, this works out to $$\C_2=4 Q^2\left(H+1\right)\,C_5\$$.

## Summary

They chose to set up their own variables. For example, assigning $$\\alpha=\frac1{Q}\$$, which makes $$\C_2=\frac{4}{\alpha^2} \left(H+1\right)\,C_5\$$. Just as they wrote in your question. (I would not have bothered to define $$\\alpha\$$. It's not needed here and it is less confusing to just use $$\Q\$$.) But it will all work out the same, when all is said and done.

\begin{align*} C_2&=4 Q^2\left(H+1\right)\,C_5\\\\ R_4&=\frac1{2 Q\,\omega_{_0}\,C_5}\\\\ R_1&=\frac1{H} \,R_4\\\\ R_3&=\frac1{H+1}\,R_4 \end{align*}

(Don't forget $$\C_2\$$ and $$\R_4\$$ may be multiplied together to get a $$\\tau=C_2\,R_4=\frac{H+1}{\zeta\,\omega_{_0}}\$$. So the new form of the denominator term, $$\\frac1{C_2\,R_4}\left[2\left(H+1\right)\right]s\$$, expands to $$\2\zeta\,\omega_{_0}s\$$. As must be.)

• Thank you very much! I was able to follow and understand all the steps. This seems like a neat algorithm for the design of this filters. Most of the time that I have tried to develop the process myself I always end up with some sort of design constraints due to square roots or minus signs that appear and make the values impossible. From your experience do you see any shortcomings of this design or design constraints that might appear. The obvious one for me is that high gain and/or high Q lead to a high value spread for the components. But it never makes the design theoretically impossible. May 11, 2023 at 17:51
• @GrangerObliviate I think you summarized the issues well. Gain impacts value spread, just as you say. For example, going from a gain of 1 to a gain of 2 may cause the capacitor value spread to go from 4 to 6 (given zeta=sqrt(2)/2 I think.) The process shown could be adapted/modified, I suppose. But I certainly could see how they developed their process. (I'd never asked myself the question you did. But when you asked it, just the asking was sufficient where I could almost immediately see how their mind worked to get there.) Reminds me... question was good. +1 May 11, 2023 at 18:17