Sensitivity of 10; ωo of Sallen-Key LPF

Meaning of $S^y_x = 10$

$$S^y_x = \frac{\Delta y/y}{\Delta x/x} = 10 \;\Rightarrow\; \frac{\Delta y}{y} \approx 10\frac{\Delta x}{x}$$

A 1% increase in $x$ causes about a 10% increase in $y$ (same direction, since the sign is positive). $y$ is very sensitive to $x$: the element $x$ must have a tight tolerance, or the circuit should be changed to a lower-sensitivity design.

Sensitivity of $\omega_o$ of the Sallen-Key low-pass filter

Sallen-Key low-pass biquad (gain $K = 1 + R_B/R_A$):

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

Its transfer function is

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

so

$$\omega_o = \frac{1}{\sqrt{R_1R_2C_1C_2}} = R_1^{-1/2}R_2^{-1/2}C_1^{-1/2}C_2^{-1/2}$$

Sensitivity with respect to $R_1$:

$$\begin{aligned} \frac{\partial\omega_o}{\partial R_1} &= -\frac12 R_1^{-3/2}(R_2C_1C_2)^{-1/2} = -\frac{\omega_o}{2R_1} \\ S^{\omega_o}_{R_1} &= \frac{R_1}{\omega_o}\cdot\left(-\frac{\omega_o}{2R_1}\right) = -\frac12 \end{aligned}$$

In the same way:

$$S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12$$

$\omega_o$ does not contain $K$, $R_A$ or $R_B$, so

$$S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0$$

So a 1% increase in any one R or C lowers $\omega_o$ by about 0.5%; if all four increase by 1%, $\omega_o$ falls by about 2%. (The Q of the Sallen-Key circuit is more sensitive: for the equal-element design $S^Q_K = 3Q - 1$, which grows with Q.)