Sensitivity of ωo and Q in Tow-Thomas LPF

Tow-Thomas low-pass biquad: OA1 is a lossy integrator ($R_2 \parallel C_1$ in feedback, inputs $V_{in}$ through $R_3$ and $V_3$ through $R_1$), OA2 is an integrator ($R_4$, $C_2$), and OA3 is a unity-gain inverter ($V_3 = -V_2$) built with two equal resistors $r$ (the gain stays $-1$ when both change together).

        +------R2------+
        +------||------+
        |      C1      |        +--||--+
Vin-R3--+--|-\         |        |  C2  |
        |  |  >--------+--R4----+-|-\  |
        |  +|/  V1     (OA1)    | |  >-+-- V2 (LP)
        | GND                GND+|/    |
        |                     (OA2)    |
        +--R1-- V3 --[inverter -1]-----+
                         (OA3)

Node equations (ideal op-amps):

$$\begin{aligned} V_1\left(\frac{1}{R_2} + sC_1\right) &= -\frac{V_{in}}{R_3} - \frac{V_3}{R_1} \\ V_2 &= -\frac{V_1}{sR_4C_2}, \qquad V_3 = -V_2 \end{aligned}$$

Solving:

$$T(s) = \frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_2C_1} + \dfrac{1}{R_1R_4C_1C_2}}$$

So

$$\omega_o = \frac{1}{\sqrt{R_1R_4C_1C_2}} = R_1^{-1/2}R_4^{-1/2}C_1^{-1/2}C_2^{-1/2}, \qquad Q = \omega_o R_2C_1 = R_2\sqrt{\frac{C_1}{R_1R_4C_2}}$$

Sensitivity of $\omega_o$: $\omega_o = R_1^{-1/2}R_4^{-1/2}C_1^{-1/2}C_2^{-1/2}$, so using $S^{x^n}_x = n$:

$$S^{\omega_o}_{R_1} = S^{\omega_o}_{R_4} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12, \qquad S^{\omega_o}_{R_2} = S^{\omega_o}_{R_3} = S^{\omega_o}_{r} = 0$$

Example of one step: $S^{\omega_o}_{R_1} = \dfrac{R_1}{\omega_o}\dfrac{\partial\omega_o}{\partial R_1} = \dfrac{R_1}{\omega_o}\left(-\dfrac{1}{2}\dfrac{\omega_o}{R_1}\right) = -\dfrac12$.

Check: sum over resistors $= -1$, sum over capacitors $= -1$, as expected for a frequency.

Sensitivity of $Q$: $Q = R_2\,C_1^{1/2}\,R_1^{-1/2}R_4^{-1/2}C_2^{-1/2}$, so

$$S^{Q}_{R_2} = 1, \quad S^{Q}_{C_1} = \frac12, \quad S^{Q}_{R_1} = S^{Q}_{R_4} = S^{Q}_{C_2} = -\frac12, \quad S^Q_{R_3} = S^Q_r = 0$$

For example, $S^Q_{R_2} = \dfrac{R_2}{Q}\cdot\dfrac{\partial Q}{\partial R_2} = \dfrac{R_2}{Q}\cdot\dfrac{Q}{R_2} = 1$.

Check: sum over resistors $= 1 - \frac12 - \frac12 = 0$, sum over capacitors $= \frac12 - \frac12 = 0$, as expected for a dimensionless quantity.

Summary

All sensitivities are at most 1 in magnitude and do not depend on $Q$, so the Tow-Thomas biquad is a low-sensitivity circuit; $R_2$ tunes $Q$ alone, and $R_3$ sets the gain alone.