# Appendix: Derivation of Design Equations

The design equations used in this article were derived using the techniques of modern filter synthesis for passive networks. This approach looks at the voltage transfer function in relation to the two-port network parameters. The idea is to simplify the problem so that the synthesis of the desired transfer function reduces to the problem of synthesizing a driving-point impedance, that is, an impedance for which the voltage and current are measured at the same pair of terminals. In this particular case, once the desired driving-point impedance is determined, it is synthesized using a Foster expansion. The Foster expansion is a partial-fraction expansion. Since a partial-fraction expansion represents a sum, this sum represents a series combination of impedances. The individual terms of the sum are then recognized as parallel RC networks for which the values can be obtained by inspection. The usual approach is to compute the component values in terms of normalized impedances, then impedance scale the network afterward. This simplifies the algebra. The synthesis procedure is normally able to achieve a circuit with the desired poles and zeros (providing the problem is well-formed), which only gets the desired transfer function within a multiplicative constant. Practical considerations are then applied to determine the relevant scale factors, usually from the transfer function's value at DC or at infinity. We'll see this technique at work in a moment.

The desired transfer function of the op-amp circuit takes the form:

(34) | $$\frac{{V}_{\mathit{out}}\left(s\right)}{{V}_{\mathit{in}}\left(s\right)}=\frac{A\left(s+{\omega}_{2}\right)\left(s+{\omega}_{4}\right)}{\left(s+{\omega}_{1}\right)\left(s+{\omega}_{3}\right)}$$ |

where A is a constant which we will initially ignore. We assume an ideal op-amp, so we want to synthesize a feedback network whose transfer function β(s) is the reciprocal of equation (34). This gives:

(35) | $$\beta \left(s\right)=\frac{B\left(s+{\omega}_{1}\right)\left(s+{\omega}_{3}\right)}{\left(s+{\omega}_{2}\right)\left(s+{\omega}_{4}\right)}=\frac{{V}_{2}\left(s\right)}{{V}_{1}\left(s\right)}$$ |

where the port numbering convention is given in Figure 2. We'll initially ignore the constant B as well, and concentrate only on obtaining the desired poles and zeros. The two-port z-parameters of the network are expressed as:

(36) | $$\begin{array}{c}{V}_{1}={z}_{11}{I}_{1}+{z}_{12}{I}_{2}\\ {V}_{2}={z}_{21}{I}_{1}+{z}_{22}{I}_{2}\end{array}$$ |

We consider all the components to be internal to the network, and we look for the voltage transfer function into an open-circuit load I_{2 }= 0 (since the op-amp inverting input is assumed to draw no current). Plugging the open-circuit case into equation (36) yields:

(37) | $$\frac{{V}_{2}\left(s\right)}{{V}_{1}\left(s\right)}=\frac{{z}_{21}}{{z}_{11}}$$ |

Looking at z_{21}, we see that by definition of equation (36),

(38) | $${z}_{21}={\left. \frac{{V}_{2}}{{I}_{1}}\right|}_{{I}_{2}=0}$$ |

By inspection of Figure 2, we can see that z_{21} = R_{3}. That is, z_{21} is independent of frequency. So the frequency dependence of the voltage transfer function is completely contained within z_{11}(s), a driving-point impedance. We can therefore synthesize z_{11}(s) to have the same poles and zeros as V_{out}(s)/V_{in}(s) in equation (34) to achieve the desired transfer function within a multiplicative constant. We'll choose a multiplicative constant for z_{11}(s) such that its value at infinite frequency is 1 Ohm. Thus our target for the normalized z_{11}(s) is:

(39) | $${z}_{11}\left(s\right)=\frac{\left(s+{\omega}_{2}\right)\left(s+{\omega}_{4}\right)}{\left(s+{\omega}_{1}\right)\left(s+{\omega}_{3}\right)}$$ |

We've chosen equation (39) as our driving-point impedance to synthesize, but so far we don't have any guarantee that this impedance is even realizable with simple RC networks. To determine this, we need a set of necessary and sufficient conditions for realizability of an RC driving-point impedance. The required condition is given in Weinberg [5]. The poles and zeros must alternate on the negative real axis. The lowest critical frequency (pole or zero) must be a pole and the highest a zero. The possibility of a zero at infinity must be included for this to be correct. The simplest example is a capacitor. Its impedance has a pole at the origin and a zero at infinity. By looking at equation (39), we can see that since ω_{1} < ω_{2} < ω_{3} < ω_{4}, the necessary conditions are indeed met. So equation (39) represents a "legal" RC impedance.

Now we're ready for Foster's expansion. The partial-fraction expansion of equation (39) is given by:

(40) | $${z}_{11}\left(s\right)=1+\frac{{a}_{1}}{\left(s+{\omega}_{1}\right)}+\frac{{a}_{2}}{\left(s+{\omega}_{3}\right)}$$ |

The constants a_{1} and a_{2} are the residues of z_{11}(s) at each of its poles. The residue formula gives:

(41) | $$\begin{array}{c}{a}_{1}={\left. \left(s+{\omega}_{1}\right){z}_{11}\left(s\right)\right|}_{s=-{\omega}_{1}}\\ {a}_{2}=\left(s+{\omega}_{3}\right){z}_{11}\left(s\right){|}_{s=-{\omega}_{3}}\end{array}$$ |

Evaluating these equations gives:

(42) | $$\begin{array}{c}{a}_{1}=\frac{\left({\omega}_{2}-{\omega}_{1}\right)\left({\omega}_{4}-{\omega}_{1}\right)}{\left({\omega}_{3}-{\omega}_{1}\right)}\\ {a}_{2}=\frac{\left({\omega}_{3}-{\omega}_{2}\right)\left({\omega}_{4}-{\omega}_{3}\right)}{\left({\omega}_{3}-{\omega}_{1}\right)}\end{array}$$ |

We need a physical interpretation of equation (40). It's clear that the right-hand side represents three impedances in series, one of which is a 1 Ohm resistor, but the second and third terms need explanation. Let's look only at the second term on the right-hand side of equation (40). Suppose we take its reciprocal, which is now an admittance, and call it Y(s). Then we have:

(43) | $$Y\left(s\right)=\frac{\left(s+{\omega}_{1}\right)}{{a}_{1}}$$ |

This is just a sum, representing two admittances in parallel. It is of the form sC + G, where:

(44) | $$\begin{array}{c}C=\frac{1}{{a}_{1}}\\ G=\frac{{\omega}_{1}}{{a}_{1}}\end{array}$$ |

We can now use equation (42) combined with equation (44) to get explicit formulas for the normalized component values of the network. But one minor issue remains. There is only one standalone resistor, the 1 Ohm value. We need two standalone resistors to get the desired network. The solution is to simply split the 1 Ohm resistor into two series resistors having a ratio k and a sum of 1 Ohm. But will this alter the poles and/or zeros of the circuit? To determine this, we'll examine the circuit of Figure 9 below and compute its transfer function.

The voltage transfer function of this circuit is:

(45) | $$\frac{{V}_{2}\left(s\right)}{{V}_{1}\left(s\right)}=\frac{1}{k+1}\times \frac{R}{R+Z\left(s\right)}$$ |

So it's clear that varying k while keeping R constant does not change the poles or zeros of the transfer function at all, only the scale factor. This allows breaking the 1 Ohm resistor into two series resistors having a ratio k. So one resistor has value k/(k + 1) and the other has value 1/(k + 1). We now have all the information we need to write the design equations for the normalized values of resistors and capacitors. The suffix 'n' will be placed on each one to denote normalized values. We'll associate R_{1} and C_{1} with the term involving a_{1} in equation (42). Likewise, we'll associate R_{2} and C_{2} with the a_{2} term. Combining equations (42) and (44), we get:

(46) | $${R}_{\mathrm{1n}}=\frac{\left({\omega}_{2}-{\omega}_{1}\right)\left({\omega}_{4}-{\omega}_{1}\right)}{{\omega}_{1}\left({\omega}_{3}-{\omega}_{1}\right)}$$ |

(47) | $${C}_{\mathrm{1n}}=\frac{\left({\omega}_{3}-{\omega}_{1}\right)}{\left({\omega}_{2}-{\omega}_{1}\right)\left({\omega}_{4}-{\omega}_{1}\right)}$$ |

(48) | $${R}_{\mathrm{2n}}=\frac{\left({\omega}_{3}-{\omega}_{2}\right)\left({\omega}_{4}-{\omega}_{3}\right)}{{\omega}_{3}\left({\omega}_{3}-{\omega}_{1}\right)}$$ |

(49) | $${C}_{\mathrm{2n}}=\frac{\left({\omega}_{3}-{\omega}_{1}\right)}{\left({\omega}_{3}-{\omega}_{2}\right)\left({\omega}_{4}-{\omega}_{3}\right)}$$ |

(50) | $${R}_{\mathrm{3n}}=\frac{1}{k+1}$$ |

(51) | $${R}_{\mathrm{4n}}=\frac{k}{k+1}$$ |

By comparing equations (46) through (51) with equations (6) through (11), we see that the only difference is that the resistors have been multiplied by R_{SCALE}, and the capacitors divided by R_{SCALE}. One item remains, that of finding the constant A in equation (34). By assuming an ideal op-amp and calculating the gain at infinite frequency where C_{1} and C_{2} are shorts, we see by inspection from the ratio of R_{4} to R_{3} that A = k + 1. Thus we can rewrite equation (34) as:

(52) | $$\frac{{V}_{\mathit{out}}\left(s\right)}{{V}_{\mathit{in}}\left(s\right)}=\frac{\left(k+1\right)\left(s+{\omega}_{2}\right)\left(s+{\omega}_{4}\right)}{\left(s+{\omega}_{1}\right)\left(s+{\omega}_{3}\right)}$$ |

Evaluating this expression at DC and equating the result to the DC gain A_{0}, we obtain:

(53) | $${A}_{0}=\frac{\left(k+1\right){\omega}_{2}{\omega}_{4}}{{\omega}_{1}{\omega}_{3}}$$ |

This is the same as equation (12). This completes the derivation of the design equations.