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Sensitivity of 2; centre frequency of Sallen-Key LPF

Short answer

A sensitivity of 2 means a 1% increase in x causes about a 2% increase in y in the same direction. For the Sallen-Key ωo, all four frequency-determining components (R1, R2, C1, C2) have sensitivity -1/2; RA and RB have zero sensitivity.

What do you understand when sensitivity of y with respect to x is 2. Perform sensitivity analysis for center frequency of Sallen-key low pass filter with respect to all elements presents in the circuit.

Meaning of Sxy=2S^y_x = 2

Sxy=Δy/yΔx/x=2  ⇒  Δyy≈2ΔxxS^y_x = \frac{\Delta y/y}{\Delta x/x} = 2 \;\Rightarrow\; \frac{\Delta y}{y} \approx 2\frac{\Delta x}{x}

A 1% increase in xx causes about a 2% increase in yy (same direction, twice as large). For example, if y∝x2y \propto x^2 then Sxy=2S^y_x = 2.

Sensitivity of the centre frequency of the Sallen-Key low-pass filter

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

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

Its transfer function is

T(s)=K/(R1R2C1C2)s2+s(1R1C1+1R2C1+1−KR2C2)+1R1R2C1C2T(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

ωo=1R1R2C1C2=R1−1/2R2−1/2C1−1/2C2−1/2\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 R1R_1:

∂ωo∂R1=−12R1−3/2(R2C1C2)−1/2=−ωo2R1SR1ωo=R1ωo⋅(−ωo2R1)=−12\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:

SR1ωo=SR2ωo=SC1ωo=SC2ωo=−12S^{\omega_o}_{R_1} = S^{\omega_o}_{R_2} = S^{\omega_o}_{C_1} = S^{\omega_o}_{C_2} = -\frac12

ωo\omega_o does not contain KK, RAR_A or RBR_B, so

SKωo=SRAωo=SRBωo=0S^{\omega_o}_{K} = S^{\omega_o}_{R_A} = S^{\omega_o}_{R_B} = 0
ElementSωoS^{\omega_o}
R1R_1, R2R_2−1/2 each
C1C_1, C2C_2−1/2 each
RAR_A, RBR_B (gain)0

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

✓ Answer checked by Claude Sonnet 5.5

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