Sallen-Key 4th order Butterworth LPF at 10,000 rad/s

Butterworth n = 4 (pole table): $p = -0.3827 \pm j0.9239$ and $-0.9239 \pm j0.3827$, so

$$T_N(s) = \frac{1}{(s^2 + 0.7654s + 1)(s^2 + 1.8478s + 1)}, \qquad Q_1 = \frac{1}{1.8478} = 0.5412,\quad Q_2 = \frac{1}{0.7654} = 1.3066$$

Equal-component Sallen–Key biquad

                 +--------[C]----------+
                 |                     |
 Vi o--[R]-------+--[R]---+----(+)     |
                          |       A ---+---o Vo
                         [C]  +--(-)   |
                          |   |        |
                         GND  +--[RB]--+
                              |
                             [RA]
                              |
                             GND

Equal-component Sallen–Key lowpass ($R_1 = R_2 = R$, $C_1 = C_2 = C$, amplifier gain $K = 1 + R_B/R_A$):

$$T(s) = \frac{K\omega_0^2}{s^2 + \dfrac{\omega_0}{Q}s + \omega_0^2}, \qquad \omega_0 = \frac{1}{RC}, \qquad Q = \frac{1}{3 - K} \;\Rightarrow\; K = 3 - \frac{1}{Q}$$

Element values

$\omega_0 = 10{,}000\ \text{rad/s}$, $C$ = 0.01 μF (chosen):

$$R = \frac{1}{\omega_0 C} = 10\ \text{k}\Omega$$

Overall passband gain $K_1K_2$ = 2.5748 (8.21 dB).

 Vi -->[SK 1: K=1.1522]-->[SK 2: K=2.2346]--> Vo
       R=10k, C=0.01uF      R=10k, C=0.01uF
       RA=10k, RB=1.522k    RA=10k, RB=12.35k

If unity DC gain is required, the input resistor of each section is split into a divider $R_{1a} = KR$, $R_{1b} = KR/(K-1)$: section 1: 11.52 kΩ and 75.69 kΩ; section 2: 22.35 kΩ and 18.1 kΩ.

Answer: all R = 10 kΩ and C = 0.01 μF; section 1 (Q = 0.5412): K = 1.1522, $R_A$ = 10 kΩ, $R_B$ = 1.522 kΩ; section 2 (Q = 1.3066): K = 2.2346, $R_A$ = 10 kΩ, $R_B$ = 12.35 kΩ; $\omega_0$ = 10,000 rad/s, passband gain 2.5748 (8.21 dB).