4th order Butterworth LPF with two Sallen-Key biquads (1 kHz)

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

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

Sallen–Key biquad

                 +--------[C1]---------+
                 |                     |
 Vi o--[R1]------+--[R2]--+----(+)     |
                          |       A ---+---o Vo
                         [C2]   +-(-)  |
                          |     |      |
                         GND    +------+

Unity-gain Sallen–Key lowpass biquad ($C_1$ from the middle node to the output, $C_2$ to ground):

$$T(s) = \frac{\dfrac{1}{R_1R_2C_1C_2}}{s^2 + \dfrac{s}{C_1}\left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right) + \dfrac{1}{R_1R_2C_1C_2}}$$

Normalised equal-resistor design ($R_1 = R_2 = 1$, $\omega_0 = 1$): $C_1C_2 = 1$ and $\omega_0/Q = 2/C_1$, so

$$C_1 = 2Q, \qquad C_2 = \frac{1}{2Q}$$

Scaling

$\omega_0 = 2\pi(1000) = 6283.2$ rad/s. In each section the larger capacitor $C_1$ is made 0.1 μF: $k_z = C_{1n}/(\omega_0 \times 10^{-7})$, $R_1 = R_2 = k_z$, $C_2 = C_{2n}/(k_z\omega_0)$.

 Vi -->[SK 1, Q=0.5412]-->[SK 2, Q=1.3066]--> Vo
        R1=R2=1.723k        R1=R2=4.159k
        C1=0.1uF            C1=0.1uF
        C2=85.36nF          C2=14.64nF

Both sections are unity-gain followers, so the DC gain is 1 and no gain compensation is needed. Half-power frequency 1 kHz.

Answer: section 1: $R_1 = R_2$ = 1.723 kΩ, $C_1$ = 0.1 μF, $C_2$ = 85.36 nF; section 2: $R_1 = R_2$ = 4.159 kΩ, $C_1$ = 0.1 μF, $C_2$ = 14.64 nF.