Tow-Thomas LPF, poles -1000 ± j9949.87

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.

Its low-pass transfer function (output $V_2$) is

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

Design

Step 1: $\omega_0$ and $Q$ from the poles. Poles $s = -\sigma \pm j\omega_d$ with $\sigma = 1000$, $\omega_d = 9949.87$:

$$\begin{aligned} \omega_0^2 &= \sigma^2 + \omega_d^2 = 1000^2 + 9949.87^2 = 99{,}999{,}913 \approx 10^8 \;\Rightarrow\; \omega_0 \approx 10{,}000\ \text{rad/s} \\ \frac{\omega_0}{Q} &= 2\sigma \;\Rightarrow\; Q = \frac{\omega_0}{2\sigma} = \frac{10000}{2000} = 5 \end{aligned}$$

Required transfer function (dc gain $K = 1.5$):

$$T(s) = \frac{1.5\,\omega_0^2}{s^2 + 2\sigma s + \omega_0^2} = \frac{1.5 \times 10^{8}}{s^2 + 2000\,s + 10^{8}}$$

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 $C_1 = C_2 = C = 0.01\ \mu$F.

$$\begin{aligned} R_2 = R_3 = R &= \frac{1}{\omega_0 C} = \frac{1}{10{,}000 \times 0.01\times10^{-6}} = 10\ \text{k}\Omega \\ R_1 &= QR = 5 \times 10\ \text{k}\Omega = 50\ \text{k}\Omega \\ R_4 &= \frac{R_2}{K} = \frac{10\ \text{k}\Omega}{1.5} = 6.667\ \text{k}\Omega \\ r &= 10\ \text{k}\Omega \ \text{(inverter, equal resistors)} \end{aligned}$$

Final circuit: Tow-Thomas biquad (figure above) with $C_1 = C_2 = 0.01\ \mu\text{F}$, $R_1 = 50\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$; take the output at $V_2$ (OA2 output) for a positive dc gain of 1.5.

Answer: $C = 0.01\ \mu\text{F}$, $R_1 = 50\ \text{k}\Omega$, $R_2 = R_3 = 10\ \text{k}\Omega$, $R_4 = 6.667\ \text{k}\Omega$, $r = 10\ \text{k}\Omega$ (all practical values).