Sensitivity of -7; RLC lowpass ωc and Q

Meaning of $S^x_y = -7$

$$\frac{\Delta x}{x} \approx -7\frac{\Delta y}{y}$$

A 1% increase in $y$ causes about a 7% decrease in $x$ (opposite direction, seven times larger). $x$ is highly sensitive to $y$, so $y$ must be held to a very tight tolerance.

Sensitivity of RLC low-pass filter

Take the series RLC low-pass circuit with the output across the capacitor:

Vin o--R--L--+--o Vout
             |
            === C
             |
GND o--------+--o

$$T(s) = \frac{1/(sC)}{R + sL + 1/(sC)} = \frac{\dfrac{1}{LC}}{s^2 + \dfrac{R}{L}s + \dfrac{1}{LC}}$$

Comparing with $\dfrac{\omega_o^2}{s^2 + (\omega_o/Q)s + \omega_o^2}$:

$$\omega_o = \frac{1}{\sqrt{LC}} = L^{-1/2}C^{-1/2}, \qquad Q = \frac{\omega_o L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} = R^{-1}L^{1/2}C^{-1/2}$$

Here $\omega_o$ is the cutoff (pole) frequency; for the Butterworth case $Q = 0.707$ it is also the half-power frequency.

Cutoff frequency:

$$\begin{aligned} S^{\omega_o}_L &= \frac{L}{\omega_o}\frac{\partial\omega_o}{\partial L} = \frac{L}{\omega_o}\left(-\frac{\omega_o}{2L}\right) = -\frac12 \\ S^{\omega_o}_C &= -\frac12, \qquad S^{\omega_o}_R = 0 \end{aligned}$$

Quality factor:

$$\begin{aligned} S^Q_R &= \frac{R}{Q}\frac{\partial Q}{\partial R} = \frac{R}{Q}\left(-\frac{Q}{R}\right) = -1 \\ S^Q_L &= +\frac12, \qquad S^Q_C = -\frac12 \end{aligned}$$

So a 1% rise in L or C lowers the cutoff by 0.5%, while R affects only Q: a 1% rise in R lowers Q by 1%.