Tow-Thomas LPF, poles -10000 ± j17320.51

The Tow-Thomas biquad realizes $T(s) = \dfrac{K\omega_0^2}{s^2 + (\omega_0/Q)s + \omega_0^2}$ with three op-amps (lossy integrator, integrator, inverter).

            R1 (sets Q)
          +--/\/\--+
          |   C1   |
          +---||---+
          |        |
Vin--R4---+--[OA1]-+--> V1 (band-pass)
          ^
          |  V1--R3-->[OA2 integrator, C2]--> V2 (LP)
          |  V2--r--->[OA3 inverter, r/r]---> V3=-V2
          |                                    |
          +------------------R2----------------+

All op-amp (+) inputs are grounded. OA1 is a lossy integrator ($R_1 \parallel C_1$ in feedback), OA2 an ideal integrator ($C_2$ in feedback), OA3 a unity-gain inverter.

Low-pass transfer function (output $V_2$):

$$\frac{V_2}{V_{in}} = \frac{\dfrac{1}{R_3R_4C_1C_2}}{s^2 + \dfrac{s}{R_1C_1} + \dfrac{1}{R_2R_3C_1C_2}}, \quad \omega_0 = \frac{1}{\sqrt{R_2R_3C_1C_2}},\ \frac{\omega_0}{Q} = \frac{1}{R_1C_1},\ K = \frac{R_2}{R_4}$$

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 10000$, $\omega_d = 17320.51$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 10000^2 + 17320.51^2 = 400{,}000{,}067 \approx 4\times10^8 \;\Rightarrow\; \omega_0 \approx 20{,}000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{20000}{20000} = 1 \end{aligned}$$

Required transfer function (dc gain $K = 2.5$):

$$T(s) = \frac{2.5\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{2.5 \times 4\times10^{8}}{s^2 + 20000\,s + 4\times10^{8}}$$

Design equations (choose $C_1 = C_2 = C$, $R_2 = R_3 = R$):

$$R = \frac{1}{\omega_0 C}, \qquad R_1 = QR = \frac{Q}{\omega_0 C}, \qquad R_4 = \frac{R_2}{K}, \qquad r = \text{any convenient value}$$

Step 2: Element values. Given $C_1 = C_2 = C = 0.001\ \mu$F.

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{20{,}000 \times 0.001\times10^{-6}} = 50\ \text{k}\Omega \\ R_1 &= QR = 1 \times 50\ \text{k}\Omega = 50\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{50\ \text{k}\Omega}{2.5} = 20\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.001\ \mu\text{F}$, $R_1 = 50\ \text{k}\Omega$, $R_2 = R_3 = 50\ \text{k}\Omega$, $R_4 = 20\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 2.5.

Answer: $C = 0.001\ \mu\text{F}$, $R_1 = 50\ \text{k}\Omega$, $R_2 = R_3 = 50\ \text{k}\Omega$, $R_4 = 20\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).