RC-CR transformation of Sallen-Key LPF to HPF

RC-CR transformation converts an active RC low-pass filter into a high-pass filter of the same order and $\omega_0$ by replacing every frequency-determining resistor with a capacitor and every capacitor with a resistor. In normalized form, $R_i \to C_i = 1/R_i$ and $C_j \to R_j = 1/C_j$. This has the same effect as the LP to HP transformation $s \to 1/s$, because a resistor ($R$) and a capacitor ($1/sC$) swap roles. Resistors that only set the amplifier gain ($R_A$, $R_B$) are not changed, since they do not depend on frequency.

Applying it to the Sallen-Key LPF

Sallen-Key LPF: $R_1$, $R_2$ in series path, $C_1$ feedback to output, $C_2$ to ground, gain $K$:

$$T_{LP}(s) = \frac{K/(R_1R_2C_1C_2)}{s^2 + s\left(\frac{1}{R_1C_1} + \frac{1}{R_2C_1} + \frac{1-K}{R_2C_2}\right) + \frac{1}{R_1R_2C_1C_2}}$$

After RC-CR: the series elements become capacitors $C_1$, $C_2$; the feedback element and the grounded element become resistors $R_1$ (to output) and $R_2$ (to ground).

                 R1
          +-----/\/\-----------+
          |                    |
Vin--C1---+A--C2--+B---(+)     |
                  |     OA >---+--- Vo
                 R2  +-(-)     |
                  |  |         RB
                 GND +---------+
                     |
                     RA
                     |
                    GND

Analysing this circuit gives

$$T_{HP}(s) = \frac{Ks^2}{s^2 + s\left(\frac{1}{R_2C_1} + \frac{1}{R_2C_2} + \frac{1-K}{R_1C_1}\right) + \frac{1}{R_1R_2C_1C_2}}$$

For equal components ($R_1 = R_2 = R$, $C_1 = C_2 = C$): $\omega_0 = 1/RC$, $\omega_0/Q = (3-K)/RC$ (same as the LPF) and high-frequency gain $K$. So a normalized LPF with $C = 1$ F, $R = 1\ \Omega$ becomes an HPF with $R = 1\ \Omega$, $C = 1$ F at the same $\omega_0$ and $Q$.